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For school graduates. It should be taken by those who plan to enter universities for the most promising specialties, such as information security, automation and control, nanotechnology, systems analysis and control, rocket complexes and astronautics, nuclear physics and technology, and many others.
Read the general information about the exam and start preparing. There are practically no changes compared to last year in the new version of the KIM USE 2019. The only thing is that fragments of programs written in the C language disappeared from the tasks: they were replaced with fragments written in the C ++ language. And also the opportunity to write an algorithm in natural language as an answer was removed from task number 25.
Assessment of the exam
Last year, in order to pass the Unified State Exam in computer science at least for the top three, it was enough to score 42 primary points. They were given, for example, for correctly completing the first 9 items of the test.
It is not yet known exactly how it will be in 2019: we need to wait for an official order from Rosobrnadzor on the correspondence of the primary and test scores. Most likely it will appear in December. Given that the maximum primary score for the entire test has remained the same, most likely the minimum score will not change either. We focus on these tables so far:
The structure of the exam test
Computer science is the longest exam (the USE in mathematics and literature takes the same length), the duration is 4 hours.
In 2019, the test consists of two parts, including 27 items.
- Part 1: 23 tasks (1–23) with a short answer, which is a number, a sequence of letters or numbers.
- Part 2: 4 tasks (24–27) with a detailed answer, the complete solution of the tasks is written on the answer form 2.
All tasks are connected in one way or another with a computer, but it is not allowed to use it for writing a program in the tasks of group C during the exam. In addition, the tasks do not require complex mathematical calculations and the calculator is also not allowed to use.
Preparation for the exam
- Take the exam tests online for free without registration and SMS. The presented tests are identical in complexity and structure to the real exams conducted in the corresponding years.
- Download the demo versions of the Unified State Exam in Computer Science, which will help you better prepare for the exam and take it easier. All proposed tests were developed and approved for preparation for the exam by the Federal Institute for Pedagogical Measurements (FIPI). In the same FIPI, all official options for the exam.
The tasks that you will see, most likely, will not be encountered in the exam, but there will be tasks similar to the demo ones, on the same topic or simply with different numbers.
General USE figures
| Year | Minimum USE score | Average score | Number of participants | Not passed,% | Qty 100-point |
Duration exam time, min. |
| 2009 | 36 | |||||
| 2010 | 41 | 62,74 | 62 652 | 7,2 | 90 | 240 |
| 2011 | 40 | 59,74 | 51 180 | 9,8 | 31 | 240 |
| 2012 | 40 | 60,3 | 61 453 | 11,1 | 315 | 240 |
| 2013 | 40 | 63,1 | 58 851 | 8,6 | 563 | 240 |
| 2014 | 40 | 57,1 | 235 | |||
| 2015 | 40 | 53,6 | 235 | |||
| 2016 | 40 | 235 | ||||
| 2017 | 40 | 235 | ||||
| 2018 |
SPECIFICATION
control measuring materials
unified state exam 2016
in informatics and ICT
1. Purpose of KIM USE
The Unified State Exam (hereinafter - the Unified State Exam) is a form of objective assessment of the quality of training of persons who have mastered educational programs of secondary general education, using tasks of a standardized form (control measuring materials).
The Unified State Exam is conducted in accordance with Federal Law No. 273-FZ of December 29, 2012 “On Education in the Russian Federation”.
Control and measuring materials allow to establish the level of mastering by graduates of the Federal component of the state standard of secondary (complete) general education in computer science and ICT, basic and profile levels.
The results of the unified state exam in computer science and ICT are recognized by educational organizations of secondary vocational education and educational organizations of higher professional education as the results of entrance examinations in computer science and ICT.
2. Documents defining the content of the KIM USE
3. Approaches to the selection of content, the development of the structure of the KIM USE
The content of the assignments was developed on the main topics of the course of computer science and ICT, combined into the following thematic blocks: "Information and its coding", "Modeling and computer experiment", "Numeral systems", "Logic and algorithms", "Elements of the theory of algorithms", "Programming "," Architecture of computers and computer networks "," Processing of numerical information "," Technologies for information retrieval and storage ".
The content of the examination paper covers the main content of the computer science and ICT course, its most important topics, the most significant material in them, unambiguously interpreted in most of the computer science and ICT course options taught at school.
The work contains both tasks of the basic level of complexity, testing the knowledge and skills provided for by the standard of the basic level, as well as
and tasks of increased and high levels of complexity, testing the knowledge and skills provided by the standard profile level... The number of tasks in the KIM version should, on the one hand, ensure a comprehensive check of the knowledge and skills of graduates acquired during the entire period of study in the subject, and, on the other hand, meet the criteria of complexity, sustainability of results, and measurement reliability. For this purpose, KIM uses two types of tasks: with a short answer and a detailed answer. The structure of the examination paper ensures an optimal balance of tasks different types and varieties, three levels of difficulty, testing knowledge and skills at three different levels: reproduction, application in a standard situation, application in a new situation. The content of the examination paper reflects a significant part of the content of the subject. All this ensures the validity of the test results and the reliability of the measurement.
4. The structure of the KIM USE
Each version of the examination paper consists of two parts and includes 27 tasks that differ in form and level of difficulty.
Part 1 contains 23 tasks with a short answer.
AT examination paper the following types of tasks with a short answer are proposed:
- tasks for choosing and recording one or more correct answers from the proposed list of answers;
- tasks for calculating a certain value;
- tasks to establish the correct sequence, represented as a string of characters according to a certain algorithm.
The answer to the tasks of Part 1 is given by the corresponding entry in the form of a natural number or a sequence of symbols (letters and numbers), written without spaces and other separators.
Part 2 contains 4 tasks with a detailed answer.
Part 1 contains 23 tasks of basic, advanced and high difficulty levels. This part contains tasks with a short answer, implying an independent formulation and recording of an answer in the form of a number or a sequence of characters. The tasks check the material of all thematic blocks. In part 1, 12 tasks refer to basic level, 10 tasks for the increased difficulty level, 1 task for the high difficulty level.
Part 2 contains 4 tasks, the first of which is of increased difficulty level, the other 3 tasks of high difficulty level. The tasks of this part involve writing a detailed answer in any form.
Unified State Exam solution informatics
1. Assignment. How many units are in binary notation of hexadecimal number 12F016 ?
Explanation.
Let's translate the number 12F016 to binary notation: 12F016 = 1001011110000 2 .
Let's count the number of units: there are 6.
Answer: 6.
2. Task Logic functionF is given by the expression (¬z) ∧ x ∨ x ∧ y ... Determine which column of the function's truth tableF each of the variables correspondsx, y, z.
Change 1 | Change 2 | Change 3 | Function |
Write letters in the answerx, y, z in the order in which the corresponding columns go (first - the letter corresponding to the 1st column; then - the letter corresponding to the 2nd column; then - the letter corresponding to the 3rd column). Write the letters in a row in a row; no separators are required between the letters. Example. Let the expressionx → y depending on two variablesx and y , and the truth table:
Change 1 | Change 2 | Function |
Then the 1st column corresponds to the variabley , and the 2nd column corresponds to the variablex ... In the answer you need to write:yx.
Explanation.
This expression is a disjunction of two conjunctions. We can notice that both terms have a factorx. That is, for x \u003d 0 the sum will be equal to 0. So, for the variablex only the third column is suitable.
In the eighth row of the tablex \u003d 1, and the value of the function is 0. This is possible only whenz \u003d 1, y \u003d 0, i.e. variable1 -z and variable2 isy.
Answer: zyx.
3. Task In the figure on the right, the road map of the N district is shown in the form of a graph, the table contains information about the lengths of these roads (in kilometers).
Since the table and the diagram were drawn independently of each other, the numbering settlements in the table has nothing to do with the letter designations on the graph. Determine what is the length of the road from point B to point E. In the answer, write down an integer - as it is indicated in the table.
Explanation.
Point B is the only point with five roads, which means it corresponds to P6, and point E is the only point with four roads, which means it corresponds to P4.
The length of the road from P6 to P4 is 20.
Answer: 20.
4. Task The database snippet provides information about the relationship. Based on the given data, determine how many direct descendants (i.e. children and grandchildren) A.K. Pavlenko. are mentioned in Table 1.
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OR
File name masks are used for group operations with files. A mask is a sequence of letters, numbers and other characters allowed in file names, in which the following characters can also appear:
The "?" (question mark) means exactly one arbitrary character.
The "*" (asterisk) character means any sequence of characters of arbitrary length, including "*" can specify an empty sequence.
There are 6 files in the directory:
maveric.map
maveric.mp3
taverna.mp4
revolver.mp4
vera.mp3
zveri.mp3
Below are eight masks. How many of them are there that match exactly four files from the given directory?
* ver * .mp * | *? ver? *. mp? | ? * ver * .mp? * | * v * r * ?. m? p * |
??? * ???. mp * | ??? * ???. m * | * a *. * a * | * a *. * p * |
Explanation.
From table 2 we see that Pavlenko A.K. (ID 2155) has two children, their IDs are 2302 and 3002.
Pavlenko E. A. (ID 2302) has three children, and Pavlenko I. A. (ID 3002) has two.
Thus, A.K. Pavlenko has seven direct descendants: two children and five grandchildren.
Answer: 7.
OR
Let's consider each mask:
1. The mask * ver * .mp * will select five files:
maveric.mp3
taverna.mp4
revolver.mp4
vera.mp3
zveri.mp3
2. By mask *? Ver? *. Mp? three files will be selected:
maveric.mp3
taverna.mp4
zveri.mp3
3. By mask? * Ver * .mp? * Four files will be selected:
maveric.mp3
taverna.mp4
revolver.mp4
zveri.mp3
4. By mask * v * r * ?. m? P * one file will be selected:
maveric.map
5. Three files will be selected by the mask ??? * ???. Mp *:
maveric.mp3
taverna.mp4
revolver.mp4
6. The mask ??? * ???. M * will select four files:
maveric.map
maveric.mp3
taverna.mp4
revolver.mp4
7. By mask * a *. * A * one file will be selected:
maveric.map
8. By mask * a *. * P * four files will be selected:
maveric.map
maveric.mp3
taverna.mp4
vera.mp3
That is, three masks, which correspond to exactly four files from this directory.
Answer: 3.
Answer: 7 | 3
5. Task The communication channel transmits messages containing only four letters: P, O, S, T; a binary code is used for transmission, allowing unambiguous decoding. For the letters T, O, P, the following code words are used: T: 111, O: 0, P: 100.
Specify the shortest codeword for the letter C, at which the code will allow unambiguous decoding. If there are several such codes, enter the code with the smallest numerical value.
Explanation.
The letter C cannot be encoded as 0 because 0 is already taken.
The letter C cannot be encoded as 1, since the encoding of the letter T starts with 1.
The letter C cannot be coded as 10, since the coding of the letter P starts with 10.
The letter C cannot be coded as 11, since the coding of the letter T starts with 11.
The letter C can be coded as 101, which is the smallest possible value.
Answer: 101.
6. Task The input of the algorithm is a natural number N. The algorithm constructs a new number R from it as follows.
1. A binary notation of the number N is constructed.
2. Two more digits are added to this record on the right according to the following rule:
A) all the digits of the binary notation are added, and the remainder of dividing the sum by 2 is appended to the end of the number (on the right). For example, record 11100 is converted to record 111001;
B) the same actions are performed on this record - on the right, the remainder of dividing the sum of the digits by 2 is added.
The record thus obtained (it contains two digits more than the record of the original number N) is a binary record of the desired number R.
Indicate the smallest number N for which the result of the algorithm is more than 125. In your answer, write this number in decimal notation.
OR
The executor Calculator has two commands, which are assigned numbers:
1.add 2,
2. multiply by 5.
Performing the first of them, the Calculator adds 2 to the number on the screen, and performing the second, multiplies it by 5.
For example, program 2121 is program
multiply by 5,
add 2,
multiply by 5,
add 2,
which converts 1 to 37.
Write down the order of the instructions in a program that converts 2 to 24 and contains no more than four instructions. Use only team numbers.
Explanation.
This algorithm assigns at the end of the number or 10, if initially there was an odd number of ones in its binary notation, or 00 if it was even.
126 10 = 1111110 2 may result from the operation of the algorithm from the number 111112 .
11111 2 = 31 10 .
Answer: 31.
OR
Let's solve the problem from the opposite, and then write the received commands from right to left.
If the number is not divisible by 5, then received through command 1, if divisible, then through command 2.
22 + 2 \u003d 24 (team 1)
20 + 2 \u003d 22 (team 1)
4 * 5 \u003d 20 (command 2)
2 + 2 \u003d 4 (team 1)
Answer: 1211.
Answer: 31 | 1211
7. Assignment. A fragment of a spreadsheet is given. The formula was copied from cell E4 to cell D3. When copying, the addresses of the cells in the formula were automatically changed. What is the numeric value of the formula in cell D3?
\u003d $ B2 * C $ 3 |
Note: the $ sign denotes absolute addressing.
OR
A fragment of a spreadsheet is given.
\u003d (A1-3) / (B1-1) | \u003d (A1-3) / (C1-5) | C1 / (A1 - 3) |
What integer number must be written in cell A1 so that the diagram built from the values \u200b\u200bof the cells in the range A2: C2 matches the picture? It is known that all values \u200b\u200bof cells from the considered range are non-negative.
Explanation.
The formula, when copied to cell D3, changed to \u003d $ B1 * B $ 3.
B1 * B3 \u003d 4 * 2 \u003d 8.
Answer: 8.
OR
Substitute the values \u200b\u200bB1 and C1 into the formulas A2: C2:
A2 \u003d (A1-3) / 5
B2 \u003d (A1-3) / 5
C2 \u003d 10 / (A1-3)
Since A2 \u003d B2, then C2 \u003d 2 * A2 \u003d 2 * B2
Let's substitute:
10 / (A1-3) \u003d 2 * (A1-3) / 5
A1 - 3 \u003d 5
A1 \u003d 8.
Answer: 8.
8. Task Write down the number that will be printed as a result of the next program. For your convenience, the program is presented in five programming languages.
BASIC | Python |
DIM S, N AS INTEGER S \u003d 0 N \u003d 0 WHILE S S \u003d S + 8 N \u003d N + 2 WEND PRINT N | s \u003d 0 n \u003d 0 while s s \u003d s + 8 n \u003d n + 2 print (n) |
Algorithmic language | Pascal |
alg early integer n, s n: \u003d 0 s: \u003d 0 nts bye s s: \u003d s + 8 n: \u003d n + 2 kts pin n con | var s, n: integer; begin s: \u003d 0; n: \u003d 0; while s begin s: \u003d s + 8; n: \u003d n + 2 end; writeln (n) end. |
Si |
|
#include int main () (int s \u003d 0, n \u003d 0; while (s printf ("% d \\ n", n); return 0; |
|
Explanation.
While loop runs as long as condition s is true
Answer: 28.
9. Assignment. What is the minimum amount of memory (in KB) that must be reserved to store any 64 × 64 pixel bitmap image, provided the image can use 256 different colors? In the answer, write down only a whole number, you do not need to write a unit of measurement.
OR
The piece of music was recorded in mono format, digitized and saved as a file without using data compression. The size of the received file is 24 MB. Then the same piece of music was re-recorded in stereo (two-channel recording) and digitized with a resolution 4 times higher and a sampling rate of 1.5 times less than the first time. No data compression was performed. Specify the size of the rewritten file in MB. In the answer, write down only a whole number, you do not need to write a unit of measurement.
Explanation.
One pixel is encoded with 8 bits of memory.
Total 64 * 64 \u003d 2 12 pixels.
Image memory size 212 * 8 \u003d 2 15 bits \u003d 2 12 bytes \u003d 4 KB.
Answer: 4.
OR
When recording the same file in stereo format, its volume doubles. 24 * 2 \u003d 48
When its resolution is increased by 4 times, its volume also increases by 4 times. 48 * 4 \u003d 192
With a decrease in the sampling frequency by 1.5 times, its volume decreases by 1.5 times. 192 / 1.5 \u003d 128.
Answer: 128.
Answer: 4 | 128
10. Task Igor compiles a table of codewords for message transmission, each message has its own codeword. As code words, Igor uses 5-letter words, which contain only the letters P, I, P, and the letter P appears exactly 1 time. Each of the other valid letters can appear in the codeword any number of times or not at all. How many different codewords can Igor use?
Explanation.
Igor can make 24 words putting the letter P in the first place. Similarly, you can put it in second, third, fourth and fifth place. We get 5 * 24 \u003d 80 words.
Answer: 80.
11. Task Below, in five programming languages, two recursive functions (procedures) are written: F and G.
BASIC | Python |
DECLARE SUB F (n) DECLARE SUB G (n) SUB F (n) IF n\u003e 0 THEN G (n - 1) END SUB SUB G (n) PRINT "*" IF n\u003e 1 THEN F (n - 3) END SUB | def F (n): If n\u003e 0: G (n - 1) def G (n): Print ("*") If n\u003e 1: F (n - 3) |
Algorithmic language | Pascal |
alg F (integer n) early If n\u003e 0 then G (n - 1) All con alg G (integer n) early Output "*" If n\u003e 1 then F (n - 3) All con | procedure F (n: integer); forward; procedure G (n: integer); forward; procedure F (n: integer); begin If n\u003e 0 then G (n - 1); end; procedure G (n: integer); begin Writeln ("*"); If n\u003e 1 then F (n - 3); end; |
Si |
|
void F (int n); void G (int n); void F (int n) ( If (n\u003e 0) G (n - 1); void G (int n) ( Printf ("*"); If (n\u003e 1) F (n - 3); |
|
How many asterisk characters will be printed on the screen when calling F (11)?
Explanation.
Let's simulate the work of the program:
F (11)
G (10): *
F (7)
G (6): *
F (3)
G (2): *
F (-1)
Answer: 3.
12. Task In TCP / IP networking terminology, a netmask refers to a binary number that indicates which part of a host's IP address refers to the address of a network, and which part to the address of the host itself on that network. Usually, the mask is written according to the same rules as the IP address - in the form of four bytes, and each byte is written as a decimal number. In this case, in the mask, first (in the most significant bits) there are ones, and then from some bit - zeros. The network address is obtained by applying bitwise conjunction to the specified host IP address and mask.
For example, if the host IP address is 231.32.255.131 and the mask is 255.255.240.0, then the network address is 231.32.240.0.
For a host with IP address 111.81.208.27, the network address is 111.81.192.0. What is the smallest possible value of the third byte from the left in the mask? Write your answer as a decimal number.
Explanation.
Let's write the third byte of the IP address and the network address in binary notation:
208 10 = 11010000 2
192 10 = 11000000 2
We see that the first two bits from the left of the mask are ones, which means that the value is the smallest, the remaining bits must be zeros. We get that the third byte from the left of the mask is equal to 110000002 = 192 10
Answer: 192.
13. Task When registering in a computer system, each user is issued a password consisting of 15 characters and containing only characters from a 12-character set: A, B, C, D, E, F, G, H, K, L, M, N. In the database data for storing information about each user is allocated the same and minimum possible integer number of bytes. In this case, character-by-character passwords are used, all characters are encoded with the same and minimum possible number of bits. In addition to the password itself, additional information is stored in the system for each user, for which an integer number of bytes is allocated; this number is the same for all users. It took 400 bytes to store information about 20 users. How many bytes are allocated to store additional information about one user? In the answer, write down only an integer - the number of bytes.
Explanation.
According to the condition, 12 letters can be used in the number. It is known that with N bits it is possible to encode 2N different variants. Since 23 4 , then 4 bits are required to write each of the 12 characters.
To store all 15 characters of the password, 4 15 \u003d 60 bits are needed, and since an integer number of bytes is used for writing, then we take the nearest not less than a multiple of eight, this number is 64 \u003d 8 8 bits (8 bytes).
Let the amount of memory allocated for additional drives bex, then:
20 * (8+ x) \u003d 400
x \u003d 12
Answer: 12.
14. Quest Artist The editor receives a string of numbers as input and converts it. The editor can execute two commands, in both commands v and w denote strings of numbers.
A) replace (v, w).
This command replaces the first occurrence of the string v from the left on the line with the string w. For example, executing the command
replace (111, 27)
converts string 05111150 to string 0527150. If there are no occurrences of the string v in the string, then executing the replace (v, w) command does not change this string.
B) found (v).
This command checks if the string v occurs in the executor line Editor. If it is encountered, the command returns the logical value "true", otherwise it returns the value "false". Line
the executor does not change.
Cycle
WHILE condition
Sequence of commands
END WHILE
Executed as long as the condition is true.
In construction
IF condition
TO team1
ELSE team2
END IF
Command1 (if condition is true) or command2 (if condition is false) is executed.
What string will result from applying the below
program to a line consisting of 68 consecutive digits 8? In response
write down the resulting line.
START
WHILE found (222) OR found (888)
IF found (222)
THEN replace (222, 8)
ELSE replace (888, 2)
END IF
END WHILE
END
Explanation.
In 68 consecutive numbers 8 there are 22 groups of three eights, which will be replaced by 22 twos and two eights will remain.
68(8) = 22(2) + 2(8)
22(2) + 2(8) = 1(2) + 9(8)
1(2) + 9(8) = 4(2)
4(2) = 1(2) + 1(8) = 28
Answer: 28.
15. Task the figure shows a diagram of the roads connecting cities A, B, C, D, D, E, F, Z, I, K, L, M.
On each road, you can only move in one direction, indicated by the arrow.
How many different routes are there from city A to city M?
Explanation.
Let's start counting the number of paths from the end of the route - from the city M. Let NX is the number of different paths from city A to city X, N is the total number of paths. You can come to the city M from L or K, therefore N \u003d NM \u003d N L + N K. (*)
Similarly:
N K \u003d N And;
N L \u003d N And;
N И \u003d N Е + N Ж + N З
N K \u003d N E \u003d 1.
Let's add more vertices:
N B \u003d N A \u003d 1;
N B \u003d N B + N A + N G \u003d 1 + 1 + 1 \u003d 3;
N E \u003d N G \u003d 1;
N Г \u003d N A \u003d 1.
Substitute in the formula (*): N \u003d NM \u003d 4 + 4 + 4 + 1 \u003d 13.
Answer: 13.
Answer: 56
16. Task Arithmetic expression value: 98 + 3 5 - 9 - recorded in base 3. How many digits "2" are in this record?
Explanation.
Let's transform the expression:
(3 2 ) 8 + 3 5 - 3 2
3 16 + 3 5 - 3 2
3 16 + 3 5 = 100...00100000
100...00100000 - 3 2 = 100...00022200
The resulting number contains three twos.
Answer: 3
17. Quest In the search engine query language, the symbol "|" is used to denote the logical operation "OR", and the symbol "&" to denote the logical operation "AND". The table shows the requests and the number of pages found on them for a certain segment of the Internet.
How many pages (in thousands) will be found by requestHomer & Odyssey & Iliad? It is assumed that all queries were executed almost simultaneously, so that the set of pages containing all the search words did not change over time
execution of requests.
Explanation.
The number of requests in this area will be denoted by Ni. Our goal is N5.
Then from the table we find that:
N5 + N6 \u003d 355,
N4 + N5 \u003d 200,
N4 + N5 + N6 \u003d 470.
From the first and second equations: N4 + 2N5 + N6 \u003d 555.
From the last equation: N5 \u003d 85.
Answer: 85
18. Task Let us denote by m & n bitwise conjunction of non-negative integersm and n ... So, for example, 14 & 5 \u003d 11102 &0101 2 = 0100 2 = 4.
For what is the smallest non-negative integerAnd the formula
x & 25 ≠ 0 → (x & 17 \u003d 0 → x & A ≠ 0)
is identically true (i.e. takes the value 1 for any non-negative integer value of the variablex)?
Explanation.
Let's introduce the notation:
(x ∈ А) ≡ A; (x ∈ P) ≡ P; (x ∈ Q) ≡ Q.
Having transformed, we get:
¬P ∨ ¬ (Q ∧ ¬A) ∨ ¬P \u003d ¬P ∨ ¬Q ∨ A.
Logical OR is true if at least one statement is true. Condition ¬P∨ ¬Q \u003d 1 the rays (−∞, 40) and (60, ∞) satisfy. Since the expression ¬P∨ ¬Q ∨ A must be identically true, expression A must be true on the segment. Its length is 20.
Answer: 20.
Answer: 8
19. Quest The program uses a one-dimensional integer array A with indices from 0 to 9. The values \u200b\u200bof the elements are 4, 7, 3, 8, 5, 0, 1, 2, 9, 6, respectively, i.e. A \u003d 4, A \u003d 7, etc.
Define the value of a variablec after executing the next snippet of this program(written below in five programming languages).
BASIC | Python |
C \u003d 0 FOR i \u003d 1 TO 9 IF A (i) C \u003d c + 1 T \u003d A (i) A (i) \u003d A (0) A (0) \u003d t ENDIF NEXT i | C \u003d 0 For i in range (1,10): If A [i] C \u003d c + 1 t \u003d A [i] A [i] \u003d A A \u003d t |
Algorithmic language | Pascal |
c: \u003d 0 nts for i from 1 to 9 if A [i] c: \u003d c + 1 t: \u003d A [i] A [i]: \u003d A A: \u003d t all kts | c: \u003d 0; for i: \u003d 1 to 9 do if A [i] begin c: \u003d c + 1; t: \u003d A [i]; A [i]: \u003d A; A: \u003d t; end; |
Si |
|
c \u003d 0; for (i \u003d 1; i if (A [i] { c ++; t \u003d A [i]; A [i] \u003d A; A \u003d t; } |
|
Explanation.
If A [i] element of the array is less than A, then the program swaps them and increases the value of the variablec by 1. The program will be executed twice, the first time swapping A and A, since 3 from becomes equal to 2.
Answer: 2.
20. TaskBelow, the algorithm is written in five programming languages. Having received the number at the entrancex, this algorithm prints the numberM... It is known thatx \u003e 100. Specify the smallest such (i.e. greater than 100) numberx, when entered, the algorithm prints 26.
BASIC | Python |
DIM X, L, M AS INTEGER INPUT X L \u003d X M \u003d 65 IF L MOD 2 \u003d 0 THEN M \u003d 52 ENDIF WHILE L M IF L\u003e M THEN L \u003d L - M ELSE M \u003d M - L ENDIF WEND PRINT M | x \u003d int (input ()) L \u003d x M \u003d 65 if L% 2 \u003d\u003d 0: M \u003d 52 while L! \u003d M: if L\u003e M: L \u003d L - M else: M \u003d M - L print (M) |
Algorithmic language | Pascal |
alg early integer x, L, M input x L: \u003d x M: \u003d 65 if mod (L, 2) \u003d 0 then M: \u003d 52 all nts bye L M if L\u003e M then L: \u003d L - M otherwise M: \u003d M - L all kts pin M con | var x, L, M: integer; begin readln (x); L: \u003d x; M: \u003d 65; if L mod 2 \u003d 0 then M: \u003d 52; while L M do if L\u003e M then L: \u003d L - M else M: \u003d M - L; writeln (M); end. |
Si |
|
#include void main () { int x, L, M; scanf ("% d", & x); L \u003d x; M \u003d 65; if (L% 2 \u003d\u003d 0) M \u003d 52; while (L! \u003d M) ( if (L\u003e M) L \u003d L - M; else M \u003d M - L; } printf ("% d", M); } |
|
Explanation.
In the body of the cycle, the numbers M and L decrease until they become equal. For 26 to be printed as a result, both numbers must be 26 at some point. Let's go from end to beginning: in the previous step, one number was 26, and the other 26 + 26 \u003d 52. One step earlier 52 + 26 \u003d 78 and 52. Before that 78 + 52 \u003d 130 and 52. That is, the smallest possible number is 130. And since the found number is even, then M will be assigned the value 52, which will lead to the required result.
Answer: 130.
21. QuestWrite in the answer the smallest value of the input variablek, in which the program produces the same response as the input valuek \u003d 10. For your convenience, the program is presented in five programming languages.
BASIC | Python |
DIM K, I AS LONG INPUT K I \u003d 1 WHILE F (I) I \u003d I + 1 WEND PRINT I FUNCTION F (N) F \u003d N * N * N END FUNCTION FUNCTION G (N) G \u003d 2 * N + 3 END FUNCTION | def f (n): return n * n * n def g (n): return 2 * n + 3 k \u003d int (input ()) i \u003d 1 while f (i) i + \u003d 1 print (i) |
Algorithmic language | Pascal |
alg early whole i, k input k i: \u003d 1 nts bye f (i) i: \u003d i + 1 kts pin i con alg int f (int n) early value: \u003d n * n * n con alg int g (int n) early value: \u003d 2 * n + 3 con | var k, i: longint; function f (n: longint): longint; begin f: \u003d n * n * n; end; function g (n: longint): longint; begin g: \u003d 2 * n + 3; end; begin readln (k); i: \u003d 1; while f (i) i: \u003d i + 1; writeln (i) end. |
Si |
|
#include long f (long n) ( return n * n * n; } long g (long n) ( return 2 * n + 3; } int main () { long k, i; scanf ("% ld", & k); i \u003d 1; while (f (i) i ++; printf ("% ld", i); return 0; } |
|
Explanation.
This program compares and and adds toi unit until... And outputs the first value of the variablei at which
For k \u003d 10, the program will print the number 3.
Let's write the inequality: from here we get that the smallest valuek = 3.
Answer: 3.
22. QuestArtist May15 converts the number on the screen. The performer has two teams, which are assigned numbers:
1. Add 1
2. Multiply by 2
The first command increases the number on the screen by 1, the second multiplies it by 2. The program for the executor May15 is a sequence of commands. How many programs exist for which, given the initial number 2, the result is the number 29, and the computation path contains the number 14 and does not contain the number 25?
The program computation path is a sequence of results
execution of all program commands. For example, for program 121 with an initial number of 7, the trajectory will consist of numbers 8, 16, 17.
Explanation.
For addition, the displacement (commutative) law is valid, which means that the order of instructions in the program does not matter for the result.
All teams increase the initial number, so the number of teams cannot exceed (30 - 21) \u003d 9. The minimum number of teams is 3.
Thus, the commands can be 3, 4, 5, 6, 7, 8, or 9. Therefore, the order of the commands does not matter, each number of commands corresponds to one set of commands that can be arranged in any order.
Let's consider all possible sets and calculate the number of options for the location of commands in them. Set 133 has 3 possible locations. Set 1223 - 12 possible locations: this is the number of permutations with repetitions (1 + 2 + 1)! / (1! · 2! · 1!)). Set 12222 - 5 options. Set 111222 - 20 possible options. Set 11123 - 20 options. Set 111113 - 6 variants, set 1111122 - 21 variants, set 11111112 - 8 variants, set 111111111 - one variant.
In total, we have 3 + 12 + 5 + 20 + 20 + 6 + 21 + 8 + 1 \u003d 96 programs.
Answer: 96.
Answer: 96.
Answer: 13
23. QuestHow many different sets of boolean variable values \u200b\u200bare therex1 , x2 , ... x9 , y1 , y2 , ... y9 that meet all of the conditions below?
(¬ (x1 ≡ y1 )) ≡ (x2 ≡ y2 )
(¬ (x2 ≡ y2 )) ≡ (x3 ≡ y3 )
…
(¬ (x8 ≡ y8 )) ≡ (x9 ≡ y9 )
The answer does not need to list all the different sets of variable valuesx1 , x2 , ... x9 , y1 , y2 , ... y9 for which this system of equalities is satisfied. As an answer, you need to indicate the number of such sets.
Explanation.
From the last equation we find that there are three possible options for the values \u200b\u200bof x8 and y8: 01, 00, 11. Let's build a tree of options for the first and second pairs of values.
Thus, we have 16 sets of variables.
Option tree for value pair 11:
We get 45 options. Thus, the system will have 45 + 16 \u003d 61 different sets of solutions.
Answer: 61.
Answer: 1024
24. QuestA positive integer not exceeding 10 is received for processing9 ... You need to write a program that displays the sum of the digits of this number less than 7. If there are no numbers less than 7, you need to display 0. The programmer wrote the program incorrectly. Below this program for your convenience is given in five programming languages.
BASIC | Python |
DIM N, DIGIT, SUM AS LONG INPUT N SUM \u003d 0 WHILE N\u003e 0 DIGIT \u003d N MOD 10 IF DIGIT SUM \u003d SUM + 1 END IF N \u003d N \\ 10 WEND PRINT DIGIT | N \u003d int (input ()) sum \u003d 0 while N\u003e 0: digit \u003d N% 10 if digit sum \u003d sum + 1 N \u003d N // 10 print (digit) |
Algorithmic language | Pascal |
alg early integer N, digit, sum input N sum: \u003d 0 nc while N\u003e 0 digit: \u003d mod (N, 10) if digit sum: \u003d sum + 1 all N: \u003d div (N, 10) kts digit output con | var N, digit, sum: longint; begin readln (N); sum: \u003d 0; while N\u003e 0 do begin digit: \u003d N mod 10; if digit sum: \u003d sum + 1; N: \u003d N div 10; end; writeln (digit) end. |
Si |
|
#include int main () { int N, digit, sum; scanf ("% d", & N); sum \u003d 0; while (N\u003e 0) { digit \u003d N% 10; if (digit sum \u003d sum + 1; N \u003d N / 10; } printf ("% d", digit); return0; } |
|
Do the following in sequence.
1. Write what this program will display when you enter the number 456.
2. Give an example of such a three-digit number, when you enter it, the program gives the correct answer.
3. Find all errors in this program (there may be one or several of them). Each error is known to affect only one line and can be corrected without changing other lines. For each error:
1) write down the line where the mistake was made;
2) indicate how to fix the error, i.e. give the correct version of the line.
It is enough to indicate the errors and how to fix them for one programming language. Please note that you need to find errors in the existing program, and not write your own, possibly using a different solution algorithm. The error correction should only affect the line in which the error is located.
Explanation.
The solution uses a Pascal program recording. It is allowed to use the program in any of four other languages.
1. The program will display the number 4.
2. An example of a number, when entered, the program gives the correct answer: 835.
Note to the reviewer. The program is not working properly due to an incorrect variable displayed on the screen and an incorrect increase in the amount. Accordingly, the program will work correctly if the most significant digit in the number (extreme left) is equal to the sum of digits less than 7.
3. There are two errors in the program.
First mistake. Incorrect increase in the amount.
Error line:
sum: \u003d sum + 1;
Correct fix:
sum: \u003d sum + digit;
Second mistake. Invalid response output to the screen.
Error line:
writeln (digit)
Correct fix:
writeln (sum)
25. QuestAn integer array of 20 elements is given. Array elements can take integer values \u200b\u200bfrom –10,000 to 10,000 inclusive. Describe in natural language or in one of the programming languages \u200b\u200ban algorithm that allows you to find and display the number of pairs of array elements in which at least one number is divisible by 3. In this problem, a pair means two consecutive array elements. For example, for an array of five elements: 6; 2; nine; –3; 6 - answer: 4.
The initial data is declared as shown below in examples for some programming languages \u200b\u200band natural language. It is prohibited to use variables not described below, but it is allowed not to use some of the described variables.
BASIC | Python |
CONST N AS INTEGER \u003d 20 DIM A (1 TO N) AS INTEGER DIM I AS INTEGER, J AS INTEGER, K AS INTEGER FOR I \u003d 1 TO N INPUT A (I) NEXT I ... END | # is also allowed # use two # integer variables j and k a \u003d n \u003d 20 for i in range (0, n): a.append (int (input ())) ... |
Algorithmic language | Pascal |
alg early int N \u003d 20 celtab a whole i, j, k nts for i from 1 to N input a [i] kts ... con | const N \u003d 20; var a: array of integer; i, j, k: integer; begin for i: \u003d 1 to N do readln (a [i]); ... end. |
Si | Natural language |
#include #define N 20 int main () ( int a [N]; int i, j, k; for (i \u003d 0; i scanf ("% d", & a [i]); ... return 0; } | We declare an array A of 20 elements. We declare integer variables I, J, K. In a loop from 1 to 20, enter the elements of array A from 1 to 20. … |
As an answer, you need to give a program fragment (or a description of the algorithm in natural language), which should be in place of the ellipsis. You can also write the solution in another programming language (indicate the name and the version of the programming language used, for example Free Pascal 2.6) or in the form of a flowchart. In this case, you must use the same input data and variables that were suggested in the condition (for example, in a pattern written in natural language).
k: \u003d k + 1
all
kts
pin k
Pascal
k: \u003d 0;
for i: \u003d 1 to N-1 do
if (a [i] mod 3 \u003d 0) or (a mod 3 \u003d 0) then
inc (k);
writeln (k);
Si
k \u003d 0;
for (i \u003d 0; i
if (a [i]% 3 \u003d\u003d 0 || a% 3 \u003d\u003d 0)
k ++;
printf ("% d", k);
Natural language
We write in the variable K the initial value equal to 0. In the loop from the first element to the penultimate element, find the remainder of dividing the current and next array elements by 3. If the first or second of the obtained remainders is 0, increase the K variable by one. After the end of the cycle, we output the value of the variable K
26. QuestTwo players, Petya and Vanya, play the following game. There are two piles of stones in front of the players. The players take turns, Petya makes the first move. In one move, the player can add one stone to one of the piles (of his choice) or double the number of stones in the pile. For example, let there be 10 stones in one pile, and 7 stones in the other; this position in the game will be denoted by (10, 7). Then, in one move, you can get any of the four positions: (11, 7), (20, 7), (10, 8), (10, 14). In order to make moves, each player has an unlimited number of stones.
The game ends at the moment when the total number of stones in the piles becomes at least 73. The player who made the last move is considered the winner, i.e. the first to get such a position that there will be 73 stones or more in the piles.
We will say that a player has a winning strategy if he can win with any opponent's moves. Describing a player's strategy means describing what move he should make in any situation that he may encounter in a different game of the opponent. For example, with the initial positions (6, 34), (7, 33), (9, 32), Petya has a winning strategy. To win, he only needs to double the number of stones in the second heap.
Exercise 1. For each of the initial positions (6, 33), (8, 32) indicate which of the players has a winning strategy. In each case, describe the winning strategy; explain why this strategy is winning, and indicate how many moves the winner might need to win with this strategy.
Task 2. For each of the initial positions (6, 32), (7, 32), (8, 31), indicate which of the players has a winning strategy. In each case, describe the winning strategy; explain why this strategy is winning, and indicate how many moves the winner might need to win with this strategy.
Task 3. For the starting position (7, 31), indicate which of the players has a winning strategy. Describe a winning strategy; explain why this strategy is winning, and indicate how many moves the winner might need to win with this strategy. Build a tree of all the games possible with the winning strategy you specified. Imagine a tree as a picture or table.
(7,31)
Total 38
(7,31+1)=(7,32)
Total 39
(7+1,32)=(8,32)
Total 40
(8+1,32)=(9,32)
Total 41
(9,32*2)=(9,64)
Total 73
(8,32+1)=(8,33)
Total 41
(8,33*2)=(8,66)
Total 74
(8*2,32)=(16,32)
Total 48
(16,32*2)=(16,64)
Total80
(8,32*2)=(8,64)
Total 72
(8,64*2)=(8,128)
Total 136
(7+1,31)=(8,31)
Total 39
(8,31+1)=(8,32)
Total 40
(8+1,32)=(9,32)
Total 41
(9,32*2)=(9,64)
Total 73
(8,32+1)=(8,33)
Total41
(8,33*2)=(8,66)
Total 74
(8*2,32)=(16,32)
Total 48
(16,32*2)=(16,64)
Total 80
(8,32*2)=(8,64)
Total 72
(8,64*2)=(8,128)
Total 136
(7*2,31)=(14,31)
Total 45
(14,31*2)=(14,62)
Total 76
(7,31*2)=(7,62)
Total 69
(7,62*2)=(7,124)
Total 131
Exercise 1. In the initial positions (6, 33), (8, 32), Vanya has a winning strategy. With the initial position (6, 33) after the first move of Petya, one of the following four positions can be obtained: (7, 33), (12, 33), (6, 34), (6, 66). Each of these positions contains less than 73 stones. Moreover, from any of these positions, Vanya can get a position containing at least 73 stones, doubling the number of stones in the second pile. For position (8, 32), after Petya's first move, one of the following four positions can be obtained: (9, 32), (16, 32), (8, 33), (8, 64). Each of these positions contains less than 73 stones. Moreover, from any of these positions, Vanya can get a position containing at least 73 stones, doubling the number of stones in the second pile. Thus, Vanya, for any move of Petit
wins on its first move.
Task 2. In the initial positions (6, 32), (7, 32) and (8, 31) Petya has a winning strategy. At the starting position (6, 32), he must get the position (6, 33) with the first move, from the starting positions (7, 32) and (8, 31). Petya after the first move should get position (8, 32). Positions (6, 33) and (8, 32) were considered in the analysis of task 1. In these positions, the player who will move second (now Petya) has a winning strategy. This strategy is described in the analysis of task 1. Thus, Petya wins with his second move in any game of Vanya.
Task 3. In the initial position (7, 31), Vanya has a winning strategy. After Petya's first move, one of four positions may appear: (8, 31), (7, 32), (14, 31) and (7, 62). In positions (14, 31) and (7, 62) Vanya can win with one move, doubling the number of stones in the second pile. Positions (8, 31) and (7, 32) were considered during the analysis of task 2. In these positions, the player who has to make a move (now Vanya) has a winning strategy. This strategy is described in the analysis of task 2. Thus, depending on the game, Petit Vanya wins on the first or second move.
27. QuestA long-term experiment is being carried out in the physics laboratory to study the Earth's gravitational field. Every minute a positive integer is transmitted to the laboratory via the communication channel - the current reading of the Sigma 2015 device. The number of transmitted numbers in the series is known and does not exceed 10,000. All numbers do not exceed 1000. The time during which the transmission takes place can be neglected.
It is necessary to calculate the "beta value" of a series of instrument readings - the minimum even product of two readings, between the transmission of which at least 6 minutes have elapsed. If such a product fails, the answer is considered to be –1.
You are offered two tasks related to this task: task A and task B. You can solve both tasks or one of them of your choice. The final grade is set as the maximum of the grades for tasks A and B. If the solution to one of the tasks is not presented, then it is considered that the grade for this task is 0 points. Task B is a complicated version of task A, it contains additional requirements for the program.
A. Write a program in any programming language to solve the problem in which the input data will be stored in an array, after which all possible pairs of elements will be checked. Indicate the version of the programming language before the program.
ALWAYS indicate that the program is a solution to PROBLEM A.
The maximum mark for completing task A is 2 points.
B. Write a program for solving the given problem, which will be efficient both in time and memory (or at least one of these characteristics).
The program is considered effective in terms of time if the operating time
the program is proportional to the number of readings obtained from the device N, i.e. with an increase in N by k times, the running time of the program should increase by no more than k times.
A program is considered memory efficient if the size of memory used in the program for storing data does not depend on the number N and does not exceed 1 kilobyte.
Before the program, indicate the version of the programming language and briefly describe the algorithm used.
ALWAYS indicate that the program is a solution to TASK B.
The maximum mark for a correct program, effective in time and memory, is 4 points.
The maximum mark for a correct program, effective in time, but ineffective in memory, is 3 points. REMINDER! Do not forget to indicate to which task each of the programs you submitted relates to.
Input data is presented as follows. The first line specifies the number N - the total number of instrument readings. It is guaranteed that N\u003e 6. Each of the next N lines contains one positive integer - the next reading of the device.
Sample input:
11
12
45
5
3
17
23
21
20
19
18
17
The program should output one number - the product described in the condition, or –1 if such a product cannot be obtained.
Sample output for the above sample input:
54
Explanation.
Task B (the solution for task A is given below, see program 4). In order for the work to be even, at least one factor must be even, therefore, when searching for suitable works, even readings of the device can be considered paired with any others, and odd readings only with even ones.
For each indication with number k, starting with k \u003d 7, consider all pairs admissible according to the conditions of the problem, in which this indication was obtained second. The minimum product of all these pairs will be obtained if the first in the pair is taken the minimum suitable reading among all received from the beginning of the reception to the reading with the number k - 6. If the next reading is even, the minimum among the previous can be any, if odd - only even.
To obtain a time-efficient solution, you need to remember the absolute minimum and minimum even readings at each time point as you enter the data, multiply each newly obtained reading by the corresponding minimum that was 6 elements earlier, and choose the minimum of all such products.
Since each current minimum reading is used after 6 more elements are entered and after that becomes unnecessary, it is sufficient to store only the last 6 minimums. To do this, you can use an array of 6 elements and cycle through it as you enter data. The size of this array does not depend on the total number of entered readings; therefore, such a solution will be effective not only in time, but also in memory. To store the absolute and even minimums, you need to use two such arrays. Below is an example of such a program written in an algorithmic language.
Example 1. An example of a correct program in an algorithmic language. The program is effective both in time and memory.
alg
early
int s \u003d 6 | required distance between readings
integer amax \u003d 1001 | more than the maximum possible reading
int N
input N
intact a | next reading
celtab mini | current minima of the last s elements
celtab minichet | even minima of the last s elements
int i
| enter the first s readings, fix the minima
whole ma; ma: \u003d amax | minimum reading
whole rush; rush: \u003d amax | minimum even reading
nts for i from 1 to s
input a
ma: \u003d imin (ma, a)
mini: \u003d ma
minicount: \u003d run
kts
integer mp \u003d amax * amax | minimum work value
whole n
nts for i from s + 1 to N
input a
if mod (a, 2) \u003d 0
then n: \u003d a * mini
otherwise if rushing
then n: \u003d a * minicount
otherwise n: \u003d amax * amax;
all
all
mp: \u003d imin (mp, n)
ma: \u003d imin (ma, a)
if mod (a, 2) \u003d 0 then rt: \u003d imin (rt, a) all
mini: \u003d ma
minicount: \u003d run
kts
if mp \u003d amax * amax then mp: \u003d - 1 all
output mp
con
Other implementations are also possible. For example, instead of cyclically filling an array, you can shift its elements each time. In the example below, not the minimums are stored and shifted, but the original values. This requires a little less memory (one array is enough instead of two), but the time-wise solution with shifts is less efficient than with circular filling. However, the running time remains proportional to N, so the maximum score for such a solution is also 4 points.
Program 2. An example of a correct program in Pascal.
The program uses shifts, but is time and memory efficient
var
N: integer;
a: array of integer; (storage of s readings)
a_: integer; (entry of the next reading)
p: integer;
i, j: integer;
begin
readln (N);
(Enter first s numbers)
for i: \u003d 1 to s do readln (a [i]);
(Entering other values, searching for the minimum product)
ma: \u003d amax; me: \u003d amax;
mp: \u003d amax * amax;
for i: \u003d s + 1 to N do begin
readln (a_);
if a
if (a mod 2 \u003d 0) and (a
if a_ mod 2 \u003d 0 then p: \u003d a_ * ma
else if me
else p: \u003d amax * amax;
if (p
(shift the elements of the auxiliary array to the left)
for j: \u003d 1 to s - 1 do
a [j]: \u003d a;
a [s]: \u003d a_
end;
if mp \u003d amax * amax then mp: \u003d - 1;
writeln (mp)
end.
If instead of a small array of fixed size (cyclic or with shifts) all the original data (or all the current minimums) are stored, the program retains time efficiency, but becomes memory inefficient, since the required memory grows proportionally to N. Below is an example of such a program in the language Pascal. Similar (and essentially similar) programs are rated no higher than 3 points.
Program 3. An example of a correct program in Pascal. Program is time efficient but memory inefficient
const s \u003d 6; (required distance between readings)
amax \u003d 1001; (more than the maximum possible reading)
var
N, p, i: integer;
ma: integer; (minimum number without s last)
me: integer; (minimum even number without s last)
mp: integer; (minimum work value)
begin
readln (N);
(Entering all instrument readings)
for i: \u003d 1 to N do readln (a [i]);
ma: \u003d amax;
me: \u003d amax;
mp: \u003d amax * amax;
for i: \u003d s + 1 to N do
begin
if a
if (a mod 2 \u003d 0) and (a
me: \u003d a;
if a [i] mod 2 \u003d 0 then p: \u003d a [i] * ma
else if me
else p: \u003d amax * amax;
if (p
end;
if mp \u003d amax * amax then mp: \u003d -1;
writeln (mp)
end.
An exhaustive solution is also possible, in which the products of all possible pairs are found and the minimum is selected from them. Below (see Program 4) is an example of such a solution. This (and similar) solution is inefficient in terms of time and memory. It is a solution to task A, but not a solution to task B. The score for such a solution is 2 points.
Program 4. An example of a correct program in Pascal. The program is ineffective neither in time nor in memory
const s \u003d 6; (required distance between readings)
var
N: integer;
a: array of integer; (all readings of the device)
mp: integer; (minimum work value)
i, j: integer;
begin
readln (N);
(Entering device values)
for i: \u003d 1 to N do
readln (a [i]);
mp: \u003d 1000 * 1000 + 1;
for i: \u003d 1 to N-s do begin
for j: \u003d i + s to N do begin
if (a [i] * a [j] mod 2 \u003d 0) and (a [i] * a [j]
then mp: \u003d a [i] * a [j]
end;
end;
if mp \u003d 1000 * 1000 + 1 then mp: \u003d -1;
writeln (mp)
K.Yu. PolyakovUnified State Exam in Informatics:
2016 and beyond ...
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
Structural changes in 2015-2016
2
Structural changes in 2015-2016
1) removal of part A
2) reducing the number of tasks
3) unification simple tasks (4, 6, 7, 9)
Goal: to leave more time to solve
difficult tasks.
4) Python language
!
K.Yu. Polyakov, 2015
Variability!
http://kpolyakov.spb.ru
Unified State Exam in Informatics: 2016 and beyond ...
3
How many units are in binary notation
hexadecimal number 12F016.
1
2
12 102
F
11112
0
1+1+4=6
Specify the smallest number whose binary notation
contains exactly three significant zeros and three ones.
Write the answer in decimal notation
1000112 = 35
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
B1: binary number system
Unified State Exam in Informatics: 2016 and beyond ...4
B1: binary number system
number 1025
1) "on the forehead" - translate ...
2) 1025 = 1024 + 1
1024 = 100000000002
1025 = 100000000012
Answer: 2
511?
511 = 512 - 1
= 10000000002 - 1 = 1111111112
Answer: 9
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
B1: binary number system
Unified State Exam in Informatics: 2016 and beyond ...5
B1: binary number system
How many units are in binary notation decimal
{!LANG-d2505f9b527362e7f2687b3173a034fc!}
1) "on the forehead" - translate ...
2) 999 = 1023 – 16 – 8
1023 = 1024 – 1 = 11111111112
{!LANG-85bd6a8a973e64783d9d99b52914f640!}
519?
519 = 512 + 7
512 = 10000000002
7 = 1112
{!LANG-45ce8f7d7bf165ec9c1cb68dfa84ad46!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-8e635e03921f14f3d4e7fbbd6446cad5!}
Unified State Exam in Informatics: 2016 and beyond ...6
{!LANG-cc3028db58d346aeff94bfaeb2aca3f5!}
{!LANG-8877b5ca81a6b7ae09af79c645105b47!}
{!LANG-d6536e4ae14534492cc01b6c1dc8127f!}
{!LANG-9a0cf587df4357aefce77a4c0ac6ee8a!}
1) 74
2) 38
3) 60
4) 47
{!LANG-5552a9d5a371b7cf283e7c9e4d69eaa8!}
{!LANG-340e5465a47367ce3ed36ceb73b4c2e3!}
{!LANG-a991ae0a6e1a50e260677f5c0def84df!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-9d2516850e128b0dd8eeca14cd6571c5!}
Unified State Exam in Informatics: 2016 and beyond ...7
{!LANG-97e5001c799ba6bbd0cc86848b48bee5!}
{!LANG-ac0c4e91162f4d210b1e61e828fc76bd!}
1
!
{!LANG-a12a20893bb11f05d3f6fd6a9de5f7e8!}
0
{!LANG-55b7e9e1ac604138f526ec7033c4c038!}
{!LANG-04d405c2a4360b7a9b42a619567cafcd!}
0
1
{!LANG-080059a2a9c2af33876794ffecbd7cdf!}
{!LANG-9bd83b577c2a50ae58334ce02b5f3838!}
{!LANG-4793ff2562d0d725bbe7a8fd98bf3294!}
{!LANG-1571ec140b19d9494ffcd1f90e43b22e!}
1
1
F
0
1
1
{!LANG-e98c90fcf4e252f9eb48a6383a268001!}
{!LANG-da5c269ec0d3d9a52311d403be7a70a5!}
{!LANG-06f2c7e73f614052c672a35b27fd233e!}
{!LANG-b327745db2b93839bc9b605f06939815!}
{!LANG-e90638d199e38a08f48f689cd7e3133b!}
{!LANG-a4df5ce05a4eec3263a6bcde8b9fbbef!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-9d2516850e128b0dd8eeca14cd6571c5!}
Unified State Exam in Informatics: 2016 and beyond ...8
{!LANG-97e5001c799ba6bbd0cc86848b48bee5!}
{!LANG-852f0da4f990fca5fae8d7845f38a835!}
{!LANG-f88a1828205e4355f30e798c5dfa1a36!}
0
0
0
0
1
1
1
1
{!LANG-e9d49917a7ff059c352c993fbfa714dc!}
0
0
1
1
0
0
1
1
K.Yu. Polyakov, 2015
{!LANG-202faa5963b5eb45abedad269afc1254!}
0
1
0
1
0
1
0
1
F
0
1
0
1
0
0
0
1
{!LANG-ef414c11845dbf3a156c1ccd8abed3aa!}
{!LANG-de1df289569dbe5296bba31f86b34e5c!}
{!LANG-4da9d126f223f639d1136bff2e8023dc!}
{!LANG-1fdfd9775674bc4e1a5ced5ff5334c21!}
{!LANG-0429c88542abfbf5a8e39dcfddb57f13!}
{!LANG-28d521bdbce4ad66e28c9dca98aa4ae0!}
{!LANG-ff856aa289ed38bb02d23f566a22b877!}
{!LANG-3cfe5bf648cd26b919ba71b489993f91!}
{!LANG-f7e0dbd164d5f69d46bb94fabb3b3308!}
http://kpolyakov.spb.ru
{!LANG-9d2516850e128b0dd8eeca14cd6571c5!}
Unified State Exam in Informatics: 2016 and beyond ...9
{!LANG-97e5001c799ba6bbd0cc86848b48bee5!}
{!LANG-473a28dcfc6b7627d3df85bb19e4888b!}
{!LANG-fdc752c293fad94b3c05d405216010ea!}
{!LANG-f88a1828205e4355f30e798c5dfa1a36!}
0
0
0
0
1
1
1
1
{!LANG-202faa5963b5eb45abedad269afc1254!}
0
0
1
1
0
0
1
1
K.Yu. Polyakov, 2015
{!LANG-e9d49917a7ff059c352c993fbfa714dc!}
0
1
0
1
0
1
0
1
F
0
0
1
0
1
1
1
1
{!LANG-f341961dd62b9ffd6045cf8787606112!}
{!LANG-b27983f04853dba6289549a5b21cfc6e!}
{!LANG-683439857e13c65befcb151fd616b888!}
{!LANG-27ea72e3f611090ea03b860a72c21707!}
{!LANG-7e45015a8af89e2bac44b31c3f202008!}
{!LANG-2f6b18d8811f9952b8e76fa7a8f1218f!}
{!LANG-9a37c5591d1e0f056949b1c4376d5c5d!}
{!LANG-af0cb12dceeec8e0170ef49c9bfc9aa5!}
{!LANG-3f90c8db2a6517423613804221db67b6!}
{!LANG-3cfe5bf648cd26b919ba71b489993f91!}
http://kpolyakov.spb.ru
{!LANG-bc09036dcaea557382c84f2cb304fb93!}
Unified State Exam in Informatics: 2016 and beyond ...10
{!LANG-6a3d290e19942392b7a1f616a6aaed1e!}
{!LANG-fe2ceb2e94309928ef7abd34cd083d6c!}
{!LANG-fe2ceb2e94309928ef7abd34cd083d6c!}
{!LANG-54350e6527296295c6a94763140b86a9!}
{!LANG-65955ddfe3ebd899bdb8f292c4841cab!}
{!LANG-f200a305244f11d7ebaeb8566d6ba0c8!}
{!LANG-fd533e622f76f812ea4e168a44580f57!}
F
{!LANG-6afe1c2d1c1be6e9cb4891b89a1d4578!}
{!LANG-54350e6527296295c6a94763140b86a9!}
4
{!LANG-65955ddfe3ebd899bdb8f292c4841cab!}
6
3
{!LANG-f200a305244f11d7ebaeb8566d6ba0c8!}
{!LANG-fd533e622f76f812ea4e168a44580f57!}
F
11
4
5
7
4
{!LANG-6afe1c2d1c1be6e9cb4891b89a1d4578!}
30
27
10
8
2
29
{!LANG-66553459b4c6334e4253dc2092c04349!}
{!LANG-e098f3fe0fbe2b49842fc79ea3c2428a!}
{!LANG-dcf7389877c9ae58b099a3210a57c8fb!}
{!LANG-ac976d277ef2c1bbe461048d548c54ff!}
{!LANG-c553959785d8b8e6fe3331d86aca70ed!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-bc09036dcaea557382c84f2cb304fb93!}
Unified State Exam in Informatics: 2016 and beyond ...11
{!LANG-6a3d290e19942392b7a1f616a6aaed1e!}
1
1
2
2
3
45
4
5
6
6
45
55
3
15 60
2
10 40
15
20 35
4
55
2
55 60 20 55
35
45
45
{!LANG-33b1ab6257124d30480927aa3e17607b!}
{!LANG-47b3920c35171cb42516042eddb6b411!}
5
2
{!LANG-568b3fe5b1727f06e056f32116e9f7fb!}
{!LANG-014f5ec97126016e42ae6140ab3ba3f1!}
K.Yu. Polyakov, 2015
{!LANG-b4a4be0a652c7df349fb8ed89102cfb2!}
2
40
7
{!LANG-a884d6932dd59f4cb6a2dc44882f1adc!}
7
10
3
4
5
{!LANG-10ab48d2f5069f963a595e413da52ab8!}
{!LANG-604142488a9fa40af23a0cbcdb9f4deb!}
{!LANG-8b7bc2b9ffdfe1f83b34f31638fe9042!}
{!LANG-7666f8df4dc0e3c2b691fd8c322463ba!}
{!LANG-6603213674bc09ba8f55cb03aeea5145!}
{!LANG-7edcbccf3a23ac799ac2311b2398e37d!}
http://kpolyakov.spb.ru
{!LANG-50ca61522c0284b08862ac540fcf826a!}
Unified State Exam in Informatics: 2016 and beyond ...12
{!LANG-4c5f0e6129c0791e57bb738e0ec88064!}
{!LANG-f6f109b2bf894ec6e7f7316a413c18cb!}
{!LANG-a9a59b1a03b0636e76e9b70fb561131f!}
{!LANG-5a47b0cb02bf498714f4671016d064f2!}
23
24
25
K.Yu. Polyakov, 2015
34
57
35
42
http://kpolyakov.spb.ru
Unified State Exam in Informatics: 2016 and beyond ...
13
{!LANG-2a6ebc811f3ce53ffe87d3ebbd0c9a9c!}
{!LANG-455d49da068dcb6b6d06c38b008da0c4!}
{!LANG-49fcf9fd118894723f28bf6bca4c2a5e!}
{!LANG-013e89306818461c570e9cf3ac807f1b!}
{!LANG-5972340caa04cb733be486907071ff16!}
{!LANG-afe08fa9ce0050214e0a5f66ad84f078!}
{!LANG-4be11eb34ab479cc0e4316857a6e3986!}
{!LANG-b4079ebd0944ffe09a8cd23a42a0d5e0!}
1
0
{!LANG-c4bf966e0ea8fbe97b40b2d7652e4298!}
{!LANG-d8a2334d9b1d635559e91bfd80395737!}
{!LANG-7ecc8fe5e68744d7510c020c83f59305!}
11
101
{!LANG-5cae00e0285152893e4f32c48bbbea17!}
K.Yu. Polyakov, 2015
0
0
110
1
1
1
0
1
{!LANG-e746b9c435afde47bfe8022d77d80835!}
http://kpolyakov.spb.ru
{!LANG-2186a04ed5064cc53f20eae1e88c0471!}
Unified State Exam in Informatics: 2016 and beyond ...14
{!LANG-cc7ffd472748cf79566f91132082c1e0!}
{!LANG-93c6df934d549c2d152a61a6d0dc98d4!}
{!LANG-57fedfae47b59b746de45a04728a82b1!}
{!LANG-83344afe7ec57bd325f4ffe57639a74b!}
{!LANG-eef8e12a7fff0530e0cbee6222867ba6!}
{!LANG-0833e0022a0d01f788ea707b8e9e81c5!}
{!LANG-2df80938ecc6813bdf9d101fdcbd980e!}
{!LANG-eff76b2808bc6ef6db2f72401c0e5d94!}
0
{!LANG-f39d5f40873eb3e155e9983c4b5d4a74!}
{!LANG-cc1cd8a1d2a0c4355eb980b64fa19c91!}
0
1
{!LANG-5c20c4abfd38e786e1e01ee753e6ca0d!}
0
1
{!LANG-47b3920c35171cb42516042eddb6b411!}
1: 001
1
{!LANG-33b1ab6257124d30480927aa3e17607b!}
{!LANG-79d2ed2f0b0143ac61ed1dd96da42db3!}
{!LANG-06d00f472ccb10b681ffe8a610c5cebb!}
1
2
4
{!LANG-b676614ef0a0aaf1b89d0a2b5a2ab6d2!}
K.Yu. Polyakov, 2015
{!LANG-0054c92a086223bdf80b917ccdacb641!}
{!LANG-cca9f11a3d5371d4c697e5940096a1c4!}
8
http://kpolyakov.spb.ru
{!LANG-b2cc674d4f0e87bf042657569f38aad3!}
Unified State Exam in Informatics: 2016 and beyond ...15
{!LANG-6a3d8a5eb29a0d247cb7fbb8bb7fa102!}
{!LANG-a740f2ec59268c8d481deebbc181b59b!}
{!LANG-eb205ecd5170e2155b641c0a3d1314d2!}
{!LANG-716774012d62384b25197866eceb86ce!}
{!LANG-375a9f94ab4bd72bcd8feeec3d7f7476!}
{!LANG-5d62c548589ce5d1e235c2b4a39c072c!}
{!LANG-29fd7e3bd688a85e2a40c52ed05ff23f!}
{!LANG-35b40d2de7db18c7f3cdb172e0431b98!}
{!LANG-1ade68d81cef30fa58a29283fd0c4df2!}
!
{!LANG-8fb8397c2e755801dd2ebfe4abffd166!}
{!LANG-bdef06d58896f5532b64ac138de3e9ff!}
{!LANG-b8c730e54bf1e8b90c8d56591db155a2!}
{!LANG-c08486ec165a3f82db0408b027736134!}
{!LANG-6abfc6d07af1a45565a51513e0f913a5!}
K.Yu. Polyakov, 2015
31
http://kpolyakov.spb.ru
{!LANG-a5b57e4453e402297128e5232416af1e!}
Unified State Exam in Informatics: 2016 and beyond ...16
{!LANG-60022a972179058b8b533b78894d333b!}
{!LANG-27d9ab567400123c69d73aa62a38abe5!}
{!LANG-1872df4719aa0925aa8e2adc614ad06a!}
{!LANG-f85eb294c8f77bd70fa8bd9f7fa85e81!}
{!LANG-eeb787b75fa12598a580598f894db8b5!}
{!LANG-faece95473cc40e498750f73c193e194!}
{!LANG-49cf20a902452319ce702e198bbf20c6!}
{!LANG-9ef736d5fd4a34efdd64998ad383d15b!}
K.Yu. Polyakov, 2015
{!LANG-2be4700c7082924904d56c117fc72b70!}
{!LANG-5ecb3214941fde83760d3417da0ce114!}
http://kpolyakov.spb.ru
{!LANG-6d43cb1bd1b2760b65f98f18abe43cf4!}
Unified State Exam in Informatics: 2016 and beyond ...17
{!LANG-8dfe965f6e0738ec577cc28abf4040e1!}
{!LANG-833848660519d8774dbb86a930ed8aed!}
{!LANG-a5cabf04f6f223de8841b16416814210!}
{!LANG-26d4b54cf63c460a0481b5294b32bbe7!}
{!LANG-e96cf80516f5b385dc9173a1dd1d76ab!}
*.*.112.*
{!LANG-6cc78a212be849878f38b0c3a23c3ede!}
*.*.64.0
{!LANG-1badfcfa74448473e768bdccb172113a!}
192
112 = 011100002
64 = 010000002
!
K.Yu. Polyakov, 2015
{!LANG-d7a5a49f0595421f91e81f938f10734a!}
http://kpolyakov.spb.ru
{!LANG-6d43cb1bd1b2760b65f98f18abe43cf4!}
Unified State Exam in Informatics: 2016 and beyond ...18
{!LANG-8dfe965f6e0738ec577cc28abf4040e1!}
{!LANG-a8488ee4277875dc12723c3187129f4b!}
{!LANG-1cb7ec1acde66617152d263a6fbc675f!}
{!LANG-9c9f99a7fff15bde94f7e137eea08cb1!}
{!LANG-b699dfb9970de7eddf38b4f2e9b06417!}
*.*.208.*
*.*.192.0
208 =
192 =
{!LANG-9184dd17fc362b8e7aa68dc403f0251e!}
{!LANG-9184dd17fc362b8e7aa68dc403f0251e!}
110100002
110000002
111000002
110000002
192
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-462c932e72e795d3593704897d9638fd!}
Unified State Exam in Informatics: 2016 and beyond ...19
{!LANG-e479f120790a19c2b0c78588aeda823e!}
{!LANG-a648f3e2c359738a607faf278611b43c!}
{!LANG-b01b723e3e03bf57fdab1335f78ce34a!}
2)
{!LANG-a57949d2ce1061741c589e946799dd88!}
{!LANG-6e86d781ec946808c2d128c4664d0524!}
{!LANG-c947a28d5898c54c677ec5c5eb623221!}
{!LANG-ad57c72dbfdbb8f11fb7a169b8d1edcb!}
{!LANG-34fe514481e5f70758d75c3975697aba!}
{!LANG-13bcb8ca13f95e27dfc9dbe3446a40a3!}
{!LANG-40f2be1d251a5a963e6a919f10b9cc4c!}
{!LANG-b72fcaf6cf628efaa2ffa726714295bb!}
{!LANG-601cf8532e3ed9cee6c6d23cb406d086!}
{!LANG-8b59a4c25cd84bfec49387e3ca2393b2!}
{!LANG-3edc220f5e824ffdcf07cc7de1454e1f!}
{!LANG-7eed5bf4c7be7b8ecf1301d0acb71bfc!}
{!LANG-13b64e15c312fa5ef9986cf7a21fc69c!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7dc9f13296c9aa6db00fc0fce58e151a!}
Unified State Exam in Informatics: 2016 and beyond ...20
{!LANG-696cad28a413c88d3c05d94ba43e5275!}
{!LANG-b0f1a333ba07c30bf119c3f700c9f2ba!}
{!LANG-3693f4b44093ad9ae1b2195349019f68!}
{!LANG-6288e7e976773a469cc48372839b8043!}
{!LANG-81ea48f1b3cd7f5282015bd308c8e854!}
{!LANG-eb8b6c84d828dd3ab1816ed052c6df37!}
{!LANG-83ec291d1b058cde2b3341403534b261!}
{!LANG-03200d845f0f655e5f85b0b83b847b97!}
888888888…8
2 2 2
8
K.Yu. Polyakov, 2015
!
{!LANG-e934bd12f9d1c221105bb4e49c8a56d9!}
{!LANG-6a9c60fd45ceb7935895d9ec003c5632!}
{!LANG-91709d38ad49a33ff0ed59687ebedaea!}
{!LANG-a8e2eed7cf6936021b849addeb74d1df!}
68
8888 28
http://kpolyakov.spb.ru
Unified State Exam in Informatics: 2016 and beyond ...
21
{!LANG-4b670c8ee9238e61457d7ce9b55dd3c7!}
{!LANG-b4a4be0a652c7df349fb8ed89102cfb2!}
{!LANG-a884d6932dd59f4cb6a2dc44882f1adc!}
{!LANG-5de6f98ba22c01844314889f27909a6b!}
{!LANG-604142488a9fa40af23a0cbcdb9f4deb!}
{!LANG-47b3920c35171cb42516042eddb6b411!}
{!LANG-6603213674bc09ba8f55cb03aeea5145!}
K.Yu. Polyakov, 2015
{!LANG-b676614ef0a0aaf1b89d0a2b5a2ab6d2!}
{!LANG-33b1ab6257124d30480927aa3e17607b!}
{!LANG-25be4b4c01d1301bb4b2d0f4fc321e38!}
{!LANG-10ab48d2f5069f963a595e413da52ab8!}
http://kpolyakov.spb.ru
{!LANG-b0139709b1edfe638eac90c8bed35adb!}
Unified State Exam in Informatics: 2016 and beyond ...22
{!LANG-5280f28c09c0c9869c57242ebd697baf!}
{!LANG-352d6bf32f2fa91207f3f1b126457b01!}
{!LANG-f6beabac0675dc14d667d657d469b49d!}
{!LANG-b4a4be0a652c7df349fb8ed89102cfb2!}
{!LANG-a884d6932dd59f4cb6a2dc44882f1adc!}
{!LANG-5de6f98ba22c01844314889f27909a6b!}
{!LANG-604142488a9fa40af23a0cbcdb9f4deb!}
{!LANG-47b3920c35171cb42516042eddb6b411!}
{!LANG-6603213674bc09ba8f55cb03aeea5145!}
K.Yu. Polyakov, 2015
{!LANG-b676614ef0a0aaf1b89d0a2b5a2ab6d2!}
{!LANG-33b1ab6257124d30480927aa3e17607b!}
{!LANG-25be4b4c01d1301bb4b2d0f4fc321e38!}
{!LANG-10ab48d2f5069f963a595e413da52ab8!}
http://kpolyakov.spb.ru
{!LANG-2286bd9295b92b1addb1308574c3c3b1!}
Unified State Exam in Informatics: 2016 and beyond ...23
{!LANG-d0074aba96f384b300127e941565c988!}
{!LANG-5b66023ec81c9e58120fce466c5a52fb!}
{!LANG-f10761a6ecd1061550e6ff65ecfdaa52!}
{!LANG-6ef23d31d3289f774353b93facb050c0!}
{!LANG-91ae224baf07b11379d885c934c74486!}
{!LANG-3aa5e4aed92fd3aab99b9bc934a62504!}
{!LANG-3aa5e4aed92fd3aab99b9bc934a62504!}
{!LANG-11093450daf8c33b77a0fed06554850e!}
{!LANG-3aa5e4aed92fd3aab99b9bc934a62504!}
{!LANG-33936f00b48836f110869372351bbaa5!}
{!LANG-3aa5e4aed92fd3aab99b9bc934a62504!}
K.Yu. Polyakov, 2015
{!LANG-8e34908a892e86ffe4de9a3d2cbdfc68!}
{!LANG-3aa5e4aed92fd3aab99b9bc934a62504!}
{!LANG-cb1824c1986557efe31d6c2c466b5f17!}
{!LANG-3aa5e4aed92fd3aab99b9bc934a62504!}
http://kpolyakov.spb.ru
{!LANG-2286bd9295b92b1addb1308574c3c3b1!}
Unified State Exam in Informatics: 2016 and beyond ...24
{!LANG-d0074aba96f384b300127e941565c988!}
{!LANG-02327969da99571f9a79b6ec4151f330!}
{!LANG-fb358672bb1246f887e62218bd19a3e6!}
{!LANG-80b74330b9492dba2fdb2ed558842ae6!}
{!LANG-c6cfa8854cf2d3f5d673d31d93ff3a03!}
= 11…100…02
{!LANG-80b74330b9492dba2fdb2ed558842ae6!}
K.Yu. Polyakov, 2015
{!LANG-c6cfa8854cf2d3f5d673d31d93ff3a03!}
http://kpolyakov.spb.ru
{!LANG-2286bd9295b92b1addb1308574c3c3b1!}
Unified State Exam in Informatics: 2016 and beyond ...25
{!LANG-d0074aba96f384b300127e941565c988!}
{!LANG-22fe6074b263d00da9b71995f85c367b!}
{!LANG-322303ca7b8de693f90c7ea29d04519d!}
{!LANG-18df258bf300a60f2d67c6c5be059373!}
= 28800 – 1 + 24400–1
= 28800 + 24400 – 21
1
4399
1 + 4399 = 4400
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-2286bd9295b92b1addb1308574c3c3b1!}
Unified State Exam in Informatics: 2016 and beyond ...27
{!LANG-d0074aba96f384b300127e941565c988!}
{!LANG-793ea0528123cb1ba1b8954e1a1e351b!}
{!LANG-2ddae341c6b8b9c6837c7fa0bcf5afd8!}
8148 = 2444
4123 = 2246
2654
17 = 16 + 1
= 24 + 2 0
2654 + 2444 – 2246 – 24 – 20
444 – 2246 – 24 – 20
2
1
444 – 2
1 + 444 – 2 = 443
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-2286bd9295b92b1addb1308574c3c3b1!}
Unified State Exam in Informatics: 2016 and beyond ...28
{!LANG-d0074aba96f384b300127e941565c988!}
{!LANG-ab3a64c1c081585f4b9e3726dc4a689f!}
{!LANG-750ea5c2b00f40063607c3b6c4e2ae39!}
9118 = 3236
27 = 33
K.Yu. Polyakov, 2015
3236 + 3123 – 33
1
{!LANG-c96297fe660e7be4f0dfd999c1a959e6!}
http://kpolyakov.spb.ru
{!LANG-2286bd9295b92b1addb1308574c3c3b1!}
Unified State Exam in Informatics: 2016 and beyond ...29
{!LANG-a9e3ba9391b113aa780a0bb2b5cdbe47!}
{!LANG-a2973dd911c5d9b88e8ba623743d29ab!}
{!LANG-958a4fc685e25c6ad8c5d7916aab53cc!}
{!LANG-e5a257b2a73eba648b57dc94b6040c3b!}
{!LANG-312ddbc708d9ee18333e7a71ce3a5f35!}
{!LANG-2a25f68df24dcf1c82cd4afb31e4d37f!}
{!LANG-1d9a5d6d2f2d7ab3198e110feb3a20ae!}
{!LANG-a2973dd911c5d9b88e8ba623743d29ab!}
{!LANG-fc5825ad03b1345975d01a060dcdfd94!}
{!LANG-54350e6527296295c6a94763140b86a9!}
{!LANG-a7002dbb584c83819f3a1fc4df3919bb!}
{!LANG-fe2ceb2e94309928ef7abd34cd083d6c!}
{!LANG-208c77eea12e4f835a8295f207d49e90!}
450
260
50
?
{!LANG-6955ecba42ec87716d424f14ec7ff290!}
{!LANG-208c77eea12e4f835a8295f207d49e90!}
450
260
50
?
{!LANG-fe2ceb2e94309928ef7abd34cd083d6c!}
{!LANG-5c5b2a62a54451bcace57838678b29e8!}
{!LANG-54350e6527296295c6a94763140b86a9!}
{!LANG-2b61ec05db88bc10373eece32e503220!}
{!LANG-1f5b7b91bf309d65e18b67063742d160!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-84a545dee5684c525a15d3cc447377a1!}
Unified State Exam in Informatics: 2016 and beyond ...30
{!LANG-1274aa3b0c69461b19fe23d3bea4f04c!}
{!LANG-accecc7d01b1e068aa33ef6a4b45f6e1!}
{!LANG-301f10dc66bba8c3eab51e91564c0399!}
{!LANG-4ae886ffce0cb7945bdee64f0d475fbb!}
{!LANG-befd1c23f39575613ea06e062552b848!}
{!LANG-a442dd72257463e66833057e91d67562!}
{!LANG-5c956c80de49aec8cafee520bafd0d83!}
{!LANG-9ef339f4963ec44f7d954ffce7ac9245!}
{!LANG-c73b64abf63bdb0741674c06d3afc3c7!}
{!LANG-c73b64abf63bdb0741674c06d3afc3c7!}
{!LANG-1a64c698af0aa63e3766b4fda2bcd0da!}
{!LANG-ce5c24dd892bc0a9e82fd4cf98f034a3!}
{!LANG-ab542986b862bc4822f93b9dfe68a35b!}
{!LANG-e10e1db60c05f3db85433b93b47e5f20!}
K.Yu. Polyakov, 2015
{!LANG-ab542986b862bc4822f93b9dfe68a35b!}
37
40
60
77
{!LANG-25be0ec232cc626df86d42367a1b035b!}
20
{!LANG-e10e1db60c05f3db85433b93b47e5f20!}
http://kpolyakov.spb.ru
{!LANG-77a2a6ce5ac0b2be60a65901cea8de49!}
Unified State Exam in Informatics: 2016 and beyond ...31
{!LANG-dcf80135bf1741f44006d419355bddad!}
{!LANG-949aa863d0b782f52d7b1cd9c6c3b43d!}
{!LANG-fa26faa2f9d06cd01c7bcbf48ce3844c!}
{!LANG-083b1dea667af698dbe28077a7ebffd3!}
{!LANG-49ed9e076dff981b01cf8fca98af3a31!}
{!LANG-26610d0ba6c9040654ec068c35aa4e67!}
{!LANG-8a3faa588c153dbde5039f355f200e32!}
{!LANG-71324b2c51b78a0abf3656a68b1112a6!}
{!LANG-26b42432f7ddb25ed4e5bf9dae00bd71!}
{!LANG-c73b64abf63bdb0741674c06d3afc3c7!}
{!LANG-ce5c24dd892bc0a9e82fd4cf98f034a3!}
{!LANG-f0901331d6c105f4c57070ef7be81043!}
K.Yu. Polyakov, 2015
= 24
http://kpolyakov.spb.ru
{!LANG-77a2a6ce5ac0b2be60a65901cea8de49!}
Unified State Exam in Informatics: 2016 and beyond ...32
{!LANG-73c535dcb22579721ef8e4614ee5b411!}
{!LANG-4f4af4dac1b06bc9512edbd4ac820206!}<>{!LANG-22569c9e117d4a12e44f62e7b5b4e89b!}<> 0))
{!LANG-d6cda7f4947086c88ea3f948567ea357!}
{!LANG-6dd579abdd99debb277db776eccf5f04!}
{!LANG-7bc321e4a79e971f2b9212785aa1492b!}
{!LANG-f134c604354556508d469f5e71b12383!}
{!LANG-38a575dacd52728f55f7115dd21b52c0!}
{!LANG-d1524791f7253de06b8e979f99b5c52c!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-77a2a6ce5ac0b2be60a65901cea8de49!}
Unified State Exam in Informatics: 2016 and beyond ...33
{!LANG-73c535dcb22579721ef8e4614ee5b411!}
{!LANG-d51b8a0ed5cfb7af6513b2573fee1125!}
{!LANG-4f4af4dac1b06bc9512edbd4ac820206!}<>{!LANG-22569c9e117d4a12e44f62e7b5b4e89b!}<> 0))
{!LANG-378939bf46f6bdc6a057d946aa1433b6!}
{!LANG-b43f0b364f503e0f6818be24d0cf5505!}
{!LANG-68c0f339af4639540a66b049f3dd6a71!}
{!LANG-65b0b486b0bac3514d0dce3f79466585!}
5 4 3 2 1 0
49 = 110001
{!LANG-de241bd33a552adef3e2197f61514177!}
{!LANG-2069dee254e22d5442a924195221be30!}
{!LANG-d67e849af16083b8754701412312c694!}
{!LANG-7411e437c3342ec5ec1d8bc0fcec49c1!}<>
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-77a2a6ce5ac0b2be60a65901cea8de49!}
Unified State Exam in Informatics: 2016 and beyond ...34
{!LANG-73c535dcb22579721ef8e4614ee5b411!}
{!LANG-d51b8a0ed5cfb7af6513b2573fee1125!}
{!LANG-4f4af4dac1b06bc9512edbd4ac820206!}<>{!LANG-22569c9e117d4a12e44f62e7b5b4e89b!}<> 0))
{!LANG-378939bf46f6bdc6a057d946aa1433b6!}
{!LANG-b43f0b364f503e0f6818be24d0cf5505!}
{!LANG-9c7e2afae0b721e9077298a35ea7cf87!}
{!LANG-2611bcd32d171b7f732e017670965699!}<>{!LANG-619585422f487d0fee1c62e370c79082!}
{!LANG-205693cf9ac57ef34d46bfa9ef1517d0!}
{!LANG-65b0b486b0bac3514d0dce3f79466585!}
5 4 3 2 1 0
33 = 100001
!
?
{!LANG-069682886a8e58d7b3d0326277489fde!}
K.Yu. Polyakov, 2015
{!LANG-cddc55e6b4fa4d2b6a250947c2c7ae2a!}
{!LANG-bd535159befa35b4a1ed7307dd94d074!}
http://kpolyakov.spb.ru
{!LANG-77a2a6ce5ac0b2be60a65901cea8de49!}
Unified State Exam in Informatics: 2016 and beyond ...35
{!LANG-73c535dcb22579721ef8e4614ee5b411!}
{!LANG-d51b8a0ed5cfb7af6513b2573fee1125!}
{!LANG-61d257ab23960d3fd29f6e10252766c8!}<>{!LANG-317518afe9f5533abeff32450cf45683!}<> 0))
{!LANG-378939bf46f6bdc6a057d946aa1433b6!}
{!LANG-b810bc8c612d08ae5881e2e3dd0833f9!}
{!LANG-6dd579abdd99debb277db776eccf5f04!}
{!LANG-acb373538706e75ede04dd35d93753e1!}
{!LANG-e7fa379ee42ba7ab556214ca6dffb1aa!}
{!LANG-f17b8a0db47be325095ac4c00f337a0d!}
{!LANG-d1524791f7253de06b8e979f99b5c52c!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-77a2a6ce5ac0b2be60a65901cea8de49!}
Unified State Exam in Informatics: 2016 and beyond ...36
{!LANG-73c535dcb22579721ef8e4614ee5b411!}
{!LANG-d51b8a0ed5cfb7af6513b2573fee1125!}
{!LANG-61d257ab23960d3fd29f6e10252766c8!}<>{!LANG-317518afe9f5533abeff32450cf45683!}<> 0))
{!LANG-378939bf46f6bdc6a057d946aa1433b6!}
{!LANG-624c92a50b28b592a803a46c7c118e4a!}
{!LANG-9c7e2afae0b721e9077298a35ea7cf87!}
{!LANG-607adea002be9113805dd7d818015ee3!}
{!LANG-bdd7e8d88e43f6086611d4aec7f22004!}
!
{!LANG-63767f80ed221851a007c6fada935840!}
{!LANG-c1c0bdc86c90cc7d98c421148ff2a1d2!}
K.Yu. Polyakov, 2015
{!LANG-609a03ce761fd4b46add276127de7d46!}
{!LANG-bf6f33675af03215ffce83360eea3159!}
{!LANG-4146851b4f361d5ed424f9cd6ce1f025!}
http://kpolyakov.spb.ru
{!LANG-77a2a6ce5ac0b2be60a65901cea8de49!}
Unified State Exam in Informatics: 2016 and beyond ...37
{!LANG-2ad7414f7d340753089ad1a7144b6228!}
{!LANG-99170b9d11c872cbb0ed3297548ad804!}
{!LANG-d26f94f27c1e11fae4361637bf6ac5b1!}
{!LANG-5be4e7cc45277ba6055350f25b381ba1!}< A[i] then begin
{!LANG-39add9e87a256916f6e21bc85b54cd99!}
{!LANG-4c74250415c14e779dcbbb9ba26690ec!}
{!LANG-d9be7d141f6462e28a820bea8bb91afe!}
{!LANG-b3231324207da34145b13984d5a25c62!}
{!LANG-ddc670d6e0f98426cf949207d70a7706!}
{!LANG-193a27f3eed6205b73a942512c1d9f79!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7e390304acb65bf90e2b26347cdf853b!}
Unified State Exam in Informatics: 2016 and beyond ...38
{!LANG-2ad7414f7d340753089ad1a7144b6228!}
1)
2)
3)
4)
5)
6)
6
9
9
9
9
9
9
9
6
7
7
7
7
7
7
7
6
6
6
6
6
2
2
2
2
2
2
2
1
1
1
5
5
5
5
5
5
5
1
1
1
1
0
0
0
0
3
3
3
3
3
3
3
0
4
4
4
4
4
4
4
0
8
8
8
8
8
8
8
0
{!LANG-9162f69b2b78a087634092540305cf76!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7e390304acb65bf90e2b26347cdf853b!}
Unified State Exam in Informatics: 2016 and beyond ...39
{!LANG-2ad7414f7d340753089ad1a7144b6228!}
{!LANG-3051ae0aeadd70e05f22a9f65187f680!}
{!LANG-99170b9d11c872cbb0ed3297548ad804!}
{!LANG-d26f94f27c1e11fae4361637bf6ac5b1!}
{!LANG-82f201603aa445bfb6d4ae0c11c929b1!}< A then begin
{!LANG-39add9e87a256916f6e21bc85b54cd99!}
{!LANG-4c74250415c14e779dcbbb9ba26690ec!}
{!LANG-b361742115a8a772d42ca57fb4929e91!}
{!LANG-d9be7d141f6462e28a820bea8bb91afe!}
{!LANG-ddc670d6e0f98426cf949207d70a7706!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
{!LANG-0e40b6e16f8286954311c1f5255d1c1c!}
4 7 3 8 5 0 1 2 9 6
4 7 3 8 5 0 1 2 9 6
4 7 3 8 5 0 1 2 9 6
K.Yu. Polyakov, 2015
{!LANG-f4ed93ed3b03b975373e957c90538fb4!}
http://kpolyakov.spb.ru
{!LANG-7e390304acb65bf90e2b26347cdf853b!}
Unified State Exam in Informatics: 2016 and beyond ...40
{!LANG-2ad7414f7d340753089ad1a7144b6228!}
{!LANG-85e0b685aa2ac2b929713f81bbae3f99!}
{!LANG-294501ac0225a013edf42e720623958b!}
{!LANG-475feb89589783999b442da522b5acc2!}
{!LANG-c1c42a38d2197293c46f2baf30f0ed35!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
{!LANG-381eda226532c8c7f2cfdcc301b5e459!}
{!LANG-790313d94e3560ce29c5291608bd1b7d!}
{!LANG-2e3a9f649bb2bb9cc69a97bd6047c224!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7e390304acb65bf90e2b26347cdf853b!}
Unified State Exam in Informatics: 2016 and beyond ...41
{!LANG-2ad7414f7d340753089ad1a7144b6228!}
{!LANG-47b4318802bd29257b6068a6613a58c5!}
{!LANG-85e0b685aa2ac2b929713f81bbae3f99!}
{!LANG-294501ac0225a013edf42e720623958b!}
{!LANG-e6365f5467d09d40d46541a2f6ed5efc!}
{!LANG-c1c42a38d2197293c46f2baf30f0ed35!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
{!LANG-e26f951fb7f583e8ae0071ab5f9ba0d4!}
{!LANG-d4fb60abe13835607606f63bb68020b1!}
{!LANG-381eda226532c8c7f2cfdcc301b5e459!}
{!LANG-790313d94e3560ce29c5291608bd1b7d!}
{!LANG-4480f3b1bc63899421aa740c68b38de8!}
1798
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7e390304acb65bf90e2b26347cdf853b!}
Unified State Exam in Informatics: 2016 and beyond ...42
{!LANG-550123f81edeab7e6c9133d75b66bf08!}
{!LANG-4d70b37f1c1cfa9d43b0a7a77632e517!}
{!LANG-4a261c0c055e3db6244d277b758c48ad!}
{!LANG-d72cdc19791d61f62b3f0c0cff34646e!}
{!LANG-1a700caa2f41e9dc91a4a9f312d94d86!}
{!LANG-f65c0ce6dc13dff85c848362e020ebdb!}
{!LANG-42df77a391c885cc0297d1bbecc62d5c!}
{!LANG-8849d6fd79fd3283313085dfc131ab1b!}
{!LANG-be6a7f2f85cc4e7338d3ba857ae250b0!}
{!LANG-2792af1984fdd176d2ea242fc468ad1f!}
33336
{!LANG-7a78a322f1d2706b40f9e1f292027a0c!}
{!LANG-db5d36252c1120b0144fd144dce83f3f!}< b then b:= y;
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
{!LANG-3a9b1ec86dccf5de838a8923ce7b4a97!}
{!LANG-592e6409a852970f0c7b45f19b5dae9b!}
!
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-9667cbc78707b357082a702dc2ed1936!}
Unified State Exam in Informatics: 2016 and beyond ...43
{!LANG-4c5b80df9b2b77c4f05ac05aefbd0020!}
{!LANG-675d37034cb88633ae6f79d5e919abc4!}
{!LANG-5f5e8f964abcc87f1694db4b72a53fbc!}
{!LANG-6fb57ceed124685c362eaac894b24472!}
{!LANG-2d69752f1a7c16b7ff8b3be136f2eaf0!}
{!LANG-d71248ce706acceae17be6e8e9dd8ac9!}
{!LANG-42df77a391c885cc0297d1bbecc62d5c!}
{!LANG-510c348e2eefb7c9cbb357cdc3db416f!}
{!LANG-1e02fff74cc01bb26e2ca7c7e6222832!}
{!LANG-e57e5e1885fe5cfde36c0e205c21c593!}
{!LANG-0774c54e64c892f35512e32eead4906f!}
{!LANG-685fd4a7caef9ca8f312bde928511e95!}
{!LANG-5a14bd8ef4fd68e5245db1c2c167c8ea!}<>{!LANG-b6c5e486dace3b89b77c232653b4f444!}
{!LANG-dde138e94a8b47e1a365934bc31b077c!}
104
{!LANG-eef9caa29ae86a30037cf67507875d18!}
{!LANG-4ca49a6817cc42d754820fe367a1771d!}
{!LANG-89f4912541529b1253b4b8c032bef073!}
{!LANG-b1b934aa629350edf83c17a8b2714223!}
{!LANG-efcf2686ca549653c67e82d338389dd3!}
{!LANG-7c6605bdb5293943e4f81d10aa917029!}
{!LANG-9c65f66a053dc6681e692adc95e756ff!}
{!LANG-17b42735f8100d47a677f8b60f41db82!}
!
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-80bdf817310457c9ece5f09873ddc904!}
Unified State Exam in Informatics: 2016 and beyond ...44
{!LANG-3ad863d4aa42a4b2c5c64d5b0129425e!}
{!LANG-2d69752f1a7c16b7ff8b3be136f2eaf0!}
{!LANG-d820e89d00d2eeaff79217fab7cb2639!}
{!LANG-eae7a8579569eee9b75ead136e92675c!}
{!LANG-c7e6dc929a5556bbeca847724217846c!}
1
10
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
…
2
12
{!LANG-6937f70e4b48712a167e30a74d9b0918!}
3
16
{!LANG-86693f16683eb4eef7ed40404868b695!}
4
22
{!LANG-28c8e037fbf06cb5a689eea3433b1ea9!}< k do
5
30
36
{!LANG-1ecb68e0b85383b41cda522d6c744370!}
{!LANG-26e81d7a66693d535fcc0dd6786129b6!}
6
40
{!LANG-2cc2d7d65fedbe4e82bf07fdb485e19f!}<= f(i)
31 … 40
10
K.Yu. Polyakov, 2015
?
{!LANG-a5c16297f00004a1de991f6e4f20c547!}
23 … 30
8
http://kpolyakov.spb.ru
{!LANG-de085effcaaf89cc469d039098832111!}
Unified State Exam in Informatics: 2016 and beyond ...45
{!LANG-3ad863d4aa42a4b2c5c64d5b0129425e!}
{!LANG-1ee59e040359649df11752e772aba508!}
{!LANG-fad840d6cb5847b83f0651457e47ecd0!}
{!LANG-07aa048bfb540b731d62ee4eacb435d6!}
{!LANG-2d69752f1a7c16b7ff8b3be136f2eaf0!}
{!LANG-8363767ba1d5f32f72175e8e56e4f3bd!}
{!LANG-c7e6dc929a5556bbeca847724217846c!}
{!LANG-09500fa2bc9253b3b5cc996ca094631c!}< k <= f(i)
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
{!LANG-32092b467c4e5f07bb9a1434b379c8fa!}< k <= i*(i-1)+10
…
{!LANG-c313067d258829d7f23ebe5eaab9a463!}< k <= i2-i+10
{!LANG-6937f70e4b48712a167e30a74d9b0918!}
{!LANG-86693f16683eb4eef7ed40404868b695!}
{!LANG-0ab88e4b005cd92317072f4cf80dae39!}< k <= 40
{!LANG-28c8e037fbf06cb5a689eea3433b1ea9!}< k do
31 … 40
{!LANG-1ecb68e0b85383b41cda522d6c744370!}
{!LANG-26e81d7a66693d535fcc0dd6786129b6!}
{!LANG-375181b79ebd2ac95918d93da463203a!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-de085effcaaf89cc469d039098832111!}
Unified State Exam in Informatics: 2016 and beyond ...46
{!LANG-3ad863d4aa42a4b2c5c64d5b0129425e!}
{!LANG-a2f01dcc64c58e1b843c78a874b915e2!}
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{!LANG-8363767ba1d5f32f72175e8e56e4f3bd!}
{!LANG-02560465cd82f8d10440df44b82c86e1!}
{!LANG-09500fa2bc9253b3b5cc996ca094631c!}< g(k) <= f(i)
{!LANG-0f1ed4560ec9e8367754faebca5f0c62!}
{!LANG-6f1f758c4dad01f7bab146eca8fce461!}< 2k+3 <= i3
{!LANG-321ade973731142efa6619f7088e5ac6!}
3 < 23 <= i3
{!LANG-53d8c4fbe1deaa2449c3a93aba24ed63!}
{!LANG-3a9d23e408406bc6e59ed6a6322433a1!}
{!LANG-174d345f4272a8bb791ab6f11392eb20!}
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{!LANG-dfa2a3980133b854b9cec188cd5b66cc!}
{!LANG-28c8e037fbf06cb5a689eea3433b1ea9!}< g(k):
8 < 2k+3 <= 27
{!LANG-acaabf1c8ddf9cd048f5eb1a55bf5cb6!}
3 … 12
{!LANG-7b56e7a97bcf466c7c1528f0030904c2!}
{!LANG-21a466803d6dcabd744c548efd8feca2!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-de085effcaaf89cc469d039098832111!}
Unified State Exam in Informatics: 2016 and beyond ...47
{!LANG-5a48ede58d825c777509370b90dc6a73!}
{!LANG-f85f7d4e03fd2c45ffc5192cffe210d0!}
{!LANG-7997cc33ecfa03dd4192c43d3b441cf6!}
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{!LANG-87dd41e5773b7e374a7c2845a1a05ac7!}
{!LANG-bf06c45f9d4d63097ef5b21b06bfeda1!}
{!LANG-7ecb13a4e28d915952ce9a4b6839544e!}
{!LANG-e47f07068ea10b0d46679d9763ab8e0e!}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
1
1
1
2
2
3
3
5
5
7
7
10
10
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
13
13
13
13
13
13
13
13
13
13
13
0
0
0
13
13
{!LANG-2bff3dae32ba8c7259b418960ce42ca4!}
K.Yu. Polyakov, 2015
{!LANG-5554299b032d649407dd3ea363a95ccc!}
http://kpolyakov.spb.ru
{!LANG-f458afc7d591202eb692821950d6756b!}
Unified State Exam in Informatics: 2016 and beyond ...48
{!LANG-385982b26cf483a634ed43ac12e98978!}
{!LANG-a3e0101a59d32142a2f71972f36aaa6d!}
{!LANG-a6a76373be3ad609eed1a0a37e8e6dbd!}
{!LANG-42df77a391c885cc0297d1bbecc62d5c!}
{!LANG-99170b9d11c872cbb0ed3297548ad804!}
{!LANG-8849d6fd79fd3283313085dfc131ab1b!}
{!LANG-a57271ff63e35793fbaeb76bf7fe6aa3!}
{!LANG-369fccca33360fc442c2485aed9fae58!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
{!LANG-eaac4449113600e006739f6b6f54917a!}
1)
2)
3)
4)
?
?
{!LANG-97c3335452246e1399d7d7bb220e3c89!}
{!LANG-550249c46a1b34a4852892f0b9f20dbc!}
{!LANG-4e92317858dc39775af22614ab85f1e0!}
{!LANG-2386148c3c527c7a5787022376b2362c!}
{!LANG-2d14dd572fdfe7fcb99c770c9e24ccc4!}
{!LANG-1c6ac5f5e19f110d6dcc4a4df83ee8fd!}
{!LANG-2300ccef2f2458cadf884272616e7c1a!}
{!LANG-17bc5625280d60a570118812d6972360!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-266cc0a6c5727ffcecab113f6278b80b!}
Unified State Exam in Informatics: 2016 and beyond ...49
{!LANG-385982b26cf483a634ed43ac12e98978!}
{!LANG-cf8ba9d8ccf0fb907ffec615d80395c8!}
{!LANG-7b6878d782be85c1f63c00cff8b932aa!}
{!LANG-5f2cc64a6c18ac19c6b9d11a313db768!}
-1
{!LANG-3da7a10c7f2c33506bdadb34ddf15ca6!}
{!LANG-fde7b7f39caecb6431597aa9510d9d9a!}
{!LANG-550249c46a1b34a4852892f0b9f20dbc!}
{!LANG-e75098b0f6fb89c57ce49c457767b640!}
{!LANG-4e92317858dc39775af22614ab85f1e0!}
{!LANG-eb7808d032a72d66d18845abd4449d9d!}
{!LANG-e7c409f5bf5ad7f1677d2463d0e62e12!}
{!LANG-8b7424b1123d8768aeb81a9d3799f01c!}
{!LANG-3122d1cd85eaebe3d47da2ba605da143!}
{!LANG-72e0c705c053627fb9a93b2116a07c23!}
{!LANG-eab35aae3c57973bb7b7c85a28f58edb!}
{!LANG-2764ba2e4ac0856144bf0355c6a41c4a!}
{!LANG-65ac7e161ced029dc7a8f04715c0290d!}
{!LANG-fac9394927e38b7f40321fb911d94fe3!}
{!LANG-0560cb5cbe126a56adfca941e94812cf!}
-1
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
{!LANG-715c020b86d6983e2f7ce7a77c4cf95e!}
{!LANG-ea08beaf1264adbd665e21cc39e33d7a!}
?
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-266cc0a6c5727ffcecab113f6278b80b!}
Unified State Exam in Informatics: 2016 and beyond ...50
{!LANG-e279b51417097f7d2c64bedf6a954d0f!}
{!LANG-bff67b9c63857cd59b95d367ab6fe02b!}
{!LANG-97ad1821766691c1a17bfe1a16d5f546!}
{!LANG-f78c4b81a745f3705935e3c2edcd5d99!}
{!LANG-878b34ebec68409ab1827a3561e2495f!}
{!LANG-2d13f1a2adbf3bd1278769fce59ffbda!}
{!LANG-e5d851562cb6f0db3b5c8b849fbe4937!}
{!LANG-8eaae3f518b96f5b2082eacf4f34d827!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
Unified State Exam in Informatics: 2016 and beyond ...
51
{!LANG-0a72ec0d4c03a772f2526cb5f576f0bc!}
{!LANG-68934d2b15a104952f7947ccb0c2be8e!}
{!LANG-70f0ae24dd4ec09ff73d0ff9e7e11c88!}
{!LANG-1521618c37265961a4a01ff55763d44c!}
{!LANG-1255eb8dfd26e0b84619dec3dcc5dbb5!}
{!LANG-2d69752f1a7c16b7ff8b3be136f2eaf0!}
{!LANG-3da7a10c7f2c33506bdadb34ddf15ca6!}
{!LANG-531c4ef9ceab4a23afa3db20b39ff1ee!}
{!LANG-5217afdc92aa122b21e200a483f81d5d!}
{!LANG-44495182e4d4d5c028d03a344fb8969c!}
{!LANG-d28dbeafeeea7f109138fd201615c469!}
{!LANG-afc279d8b7c448849caede633f474e3a!}
{!LANG-f9a2a04eef0321a4f7394d7ba9f22c8a!}
{!LANG-c3fb61a2485207e873fb6047f252c404!}
{!LANG-17b42735f8100d47a677f8b60f41db82!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7b79bfe20be4dad70a54818eeb8b6c32!}
Unified State Exam in Informatics: 2016 and beyond ...52
{!LANG-0a72ec0d4c03a772f2526cb5f576f0bc!}
{!LANG-fce80a2ba0bcb596538c49705e32eaf0!}
{!LANG-c8a2dc1a8b98084d893600f19db19378!}
{!LANG-d820e89d00d2eeaff79217fab7cb2639!}
{!LANG-0ae64857dcabbb752f93bd15c934ecf9!}
{!LANG-55ccb1204c50d3a7a7e93233aa3db599!}
{!LANG-4b457c679206c5da2b47af9c150c3d0c!}
{!LANG-26efe59c534381b434b6f3425a9a3216!}
{!LANG-4bcde4d95b26a14da61c56206953d4a0!}
{!LANG-4bcde4d95b26a14da61c56206953d4a0!}
{!LANG-7ce1f151b897f99a2bbf21637e9e6ee0!}
{!LANG-ca5fb89767c30f659c65e588a71566dd!}
{!LANG-4e3daa48dc18ded82fdadc6e72432432!}
{!LANG-5ec406c0b646e65ad749e07b96381ff5!}
{!LANG-28feea60cebfae0017f376239ae0a761!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7b79bfe20be4dad70a54818eeb8b6c32!}
Unified State Exam in Informatics: 2016 and beyond ...53
{!LANG-0a72ec0d4c03a772f2526cb5f576f0bc!}
{!LANG-c8a2dc1a8b98084d893600f19db19378!}
{!LANG-d820e89d00d2eeaff79217fab7cb2639!}
{!LANG-1aab01b3130f4519855ca67200780d5c!}
{!LANG-55037e3fd7c8e25f8ce9c488b058a179!}
{!LANG-25be0ec232cc626df86d42367a1b035b!}
{!LANG-eb83c757001785406f8b6f87dc06f29b!}
{!LANG-f5ec777ca2e8c432cf2de3b97df8371e!}
{!LANG-788c42e686f2da698da5e76ef0185c39!}
{!LANG-fd96fa9b6487c42f3ba02589cf89e9cd!}
{!LANG-ce8e5d6784204c6cdd7d700249e0542b!}
{!LANG-764cb7ce61625866ead9fc5bcff87ef9!}
K.Yu. Polyakov, 2015
!
{!LANG-b11b7279cd4fd0b39a268015a4ff4168!}
http://kpolyakov.spb.ru
{!LANG-7b79bfe20be4dad70a54818eeb8b6c32!}
Unified State Exam in Informatics: 2016 and beyond ...54
{!LANG-0a72ec0d4c03a772f2526cb5f576f0bc!}
{!LANG-f7e7cd5b7528a5d3eb87a3b91f11f3f0!}
{!LANG-d03b3c035cbe2ce1ef06004cf26344da!}
{!LANG-25be0ec232cc626df86d42367a1b035b!}
{!LANG-c51def982c274d20de429ccb45adbd70!}
{!LANG-b9d2bb21c9d39beaf46f533caf265f66!}
{!LANG-7ce1f151b897f99a2bbf21637e9e6ee0!}
{!LANG-ca5fb89767c30f659c65e588a71566dd!}
{!LANG-01cf766af61ea0713757e35127b91d42!}
{!LANG-c26371901895a18ad5f6d69037792417!}
{!LANG-5ec406c0b646e65ad749e07b96381ff5!}
{!LANG-f387cb7649940723316d23039ffaa23b!}
{!LANG-96c0aa266086a0cd1ff7390627ca3fb4!}
{!LANG-484d81b8771dbc08aefd3009b448a703!}
{!LANG-7df7529acb48936b7f1adf50d2d0448f!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7b79bfe20be4dad70a54818eeb8b6c32!}
Unified State Exam in Informatics: 2016 and beyond ...55
{!LANG-0a72ec0d4c03a772f2526cb5f576f0bc!}
{!LANG-ea6afd02767edacc0882859c421daa32!}
{!LANG-1491c4e5fea48c9b28192726c32e23f3!}
1
2
3
9
1
5
6
7
{!LANG-19f8435770c6cf15b75d13df6d3fb29b!}
0
{!LANG-d03b3c035cbe2ce1ef06004cf26344da!}
4
10
2 11
3 12
4 5
8
9
{!LANG-90a4842d714d90975916fc18e17bc794!}
10 11 12 13 14 15 16 17 18
7
6
7
8
{!LANG-eb9a9027f44f21c26515c6f4a4bfa80e!}
{!LANG-b559b03cea549350688722621d315aba!}
{!LANG-683ac7480adbab6df313794a291f2455!}
{!LANG-c26371901895a18ad5f6d69037792417!}
{!LANG-0b2abfe5147eb8601856c487a95742f7!}
{!LANG-30496c290b05a24489c8376f6cfa6589!}
{!LANG-f387cb7649940723316d23039ffaa23b!}
{!LANG-96fb5f55f64d952d3d3050ad0cc4d099!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7b79bfe20be4dad70a54818eeb8b6c32!}
Unified State Exam in Informatics: 2016 and beyond ...56
{!LANG-0a72ec0d4c03a772f2526cb5f576f0bc!}
{!LANG-cac7a99673a0a2e355b2f525d2b1b43c!}
{!LANG-d1407118f7390e994ddaa6c0c1424f3a!}
{!LANG-1f96f94ab8824d5180e9eeeff0b9320c!}
{!LANG-25be0ec232cc626df86d42367a1b035b!}
{!LANG-ff56e82face64fd565efd750563e5f31!}
{!LANG-798a599a7232a1d3b2ae6ee0f3e70c7e!}
{!LANG-9369fbffdc901a612d2e84fb04068f31!}
{!LANG-25be0ec232cc626df86d42367a1b035b!}
{!LANG-f19943fd500499d9a43dc71bee8f5efb!}
{!LANG-a4edf4930828057b0e37449decbb6035!}
K.Yu. Polyakov, 2015
{!LANG-1aab01b3130f4519855ca67200780d5c!}
{!LANG-5707cc0e7e32822ea26fa8b89b4471c0!}
http://kpolyakov.spb.ru
{!LANG-7b79bfe20be4dad70a54818eeb8b6c32!}
Unified State Exam in Informatics: 2016 and beyond ...57
{!LANG-0a72ec0d4c03a772f2526cb5f576f0bc!}
{!LANG-683ac7480adbab6df313794a291f2455!}
{!LANG-c26371901895a18ad5f6d69037792417!}
{!LANG-0b2abfe5147eb8601856c487a95742f7!}
{!LANG-ce1363e4dc406a353f29444bf2044680!}
{!LANG-cd099c7baf060d7bff0cc9aa87cecda5!}
{!LANG-30496c290b05a24489c8376f6cfa6589!}
{!LANG-581390d3fe1ef8f4b444d1dde1755277!}
{!LANG-3713fbc4b3c9a8324c6106b3c02fd49a!}
{!LANG-1e7d7dc3e7e3fe787c500675db9420df!}
{!LANG-f60286df8b99a8c6a78f843125dfca2c!}
{!LANG-c7e21e7064d49d0f516a8d4d52e844a2!}
{!LANG-40ddf8f95d039508f8ad3aa52b90c870!}
{!LANG-fe6a5ffd242086b765e30d4464c0761c!}
{!LANG-b6e8ad408b9850d0ea0f38869600405d!}
{!LANG-f60286df8b99a8c6a78f843125dfca2c!}
{!LANG-cd099c7baf060d7bff0cc9aa87cecda5!}
{!LANG-b1b934aa629350edf83c17a8b2714223!}
{!LANG-f387cb7649940723316d23039ffaa23b!}
{!LANG-96fb5f55f64d952d3d3050ad0cc4d099!}
{!LANG-fb0789ea1a06fbdd5e62b2183da16d16!}
K.Yu. Polyakov, 2015
http://kpolyakov.spb.ru
{!LANG-7b79bfe20be4dad70a54818eeb8b6c32!}
Unified State Exam in Informatics: 2016 and beyond ...58
{!LANG-36e659aa8aec5ad5f5414ae72841c12d!}
!
K.Yu. Polyakov, 2015
Variability!
http://kpolyakov.spb.ru
{!LANG-5d1574f688b4121b86a173e17703d519!}
Unified State Exam in Informatics: 2016 and beyond ...59
{!LANG-2ce0ef2f7c987338c1edf39bca1407b2!}
{!LANG-984eb1f4fa1b359c0465edb5eef9d7a6!}
{!LANG-f0388265b82f77b4ffc37e60f27a95a6!}
{!LANG-43e05e0a19b8663dc902bd57f8cf881c!}
K.Yu. Polyakov, 2015
{!LANG-0b2194193640fe33f526ad141a90f087!}
