1. Computer science textbook, typed on a computer, contains 256 pages, 40 lines on each page, 60 characters in each line. To encode characters, the KOI-8 encoding is used, in which each character is encoded with 8 bits. Determine the information volume of the textbook.
2) 200 KB
3) 600 KB
4) 1200 bytes
Explanation.
Find the number of characters in the article:
256 40 60 = 2 8 5 15 2 5 = 75 2 13 .
One character is encoded by one byte, 2 10 bytes make up 1 kilobyte, so the information volume of the article is
75 8 2 10 bytes = 600 KB.
2. The text of the story is typed on a computer. The information volume of the resulting file is 9 KB. The text occupies 6 pages, each page has the same number of lines, each line has 48 characters. All characters are represented in KOI-8 encoding, in which each character is encoded with 8 bits. Determine how many lines fit on each page.
Explanation.
Information volume of the file V = 8PSC, where P- number of pages, S-number of lines C- number of characters in a string, multiplier 8 is the informational weight of one character in bits. Where do we get:
S = V/(8PC)=9 2 10 2 3 /(8 6 48) = 32
There are 32 lines on one page.
The correct answer is number 3.
3. In one of the Unicode encodings, each character is encoded with 16 bits. Determine the size of the next sentence in the given encoding. Seven times measure cut once!
Explanation.
There are 33 characters in the sentence. Therefore, the Unicode sentence size is: 33 16 = 528 bits.
The correct answer is number 4.
4. For which of the given names is the statement false:
NOT((First letter consonant) AND(The last letter is a vowel))?
Explanation.
Convert AND to OR according to De Morgan's rules:
NOT(First letter consonant) OR NOT(Last letter is a vowel)
Let's write an equivalent statement:
(First letter is a vowel) OR(Last letter consonant)
The logical "OR" is false only when both statements are false. Let's check all the answers.
1) False, since both statements are false: q is a consonant and i is a vowel.
2) True, since the second statement is true: l is a consonant.
3) True, since both statements are true: a is a vowel and m is a consonant.
4) True, since the first statement is true: a is a vowel.
5. For which of the following names of Russian writers and poets is the statement true:
NOT (the number of vowels is even) AND NOT (the first letter is a consonant)?
1) Yesenin
2) Odoevsky
3) Tolstoy
Explanation.
The logical "AND" is true only when both statements are true. Let's check all the answers.
1) Yesenin is true, since both statements are true.
2) Odoevsky is false, because the statement "NOT (the number of vowels is even)" is false.
3) Tolstoy is false, because the statement "NOT (the first letter of a consonant)" is false.
4) Fet is false, since both statements are false.
The correct answer is number 1.
6. For which of the given values of the number X true statement :( X < 5) AND NOT (X < 4)?
Explanation.
The logical "AND" is true only when both statements are true. We write the expression in the form
(X < 5)AND (X >= 4)
And check all the answers.
1) False, because the first statement is false: 5 is less than 5.
2) False, because the second statement is false: 2 is not less than 4.
3) False, because the second statement is false: 3 is not less than 4.
4) True, since both statements are true: 4 is less than 5 and 4 is not less than 4.
The correct answer is number 4.
7. Roads were built between settlements A, B, C, D, E, the length of which (in kilometers) is given in the table:
Explanation.
From point A you can get to points B, D.
From point B you can get to points C, D.
A-D-B-C-E: route length 12 km.
A-D-C-E: route length 9 km.
A-B-D-C-E: route length 8 km.
8. Roads were built between settlements A, B, C, D, E, the length of which (in kilometers) is shown in the table:
Determine the length of the shortest path between points A and E. You can only move along the roads, the length of which is indicated in the table.
Explanation.
Find all route options from A to E and choose the shortest one.
From point A you can go to point B.
From point B you can get to points C, D, E.
From point C you can go to point E.
From point D you can go to point E.
A-B-C-E: route length 9 km.
A-B-E: route length 9 km.
A-B-D-E: route length 7 km.
The correct answer is number 3.
9. Roads were built between settlements A, B, C, D, E, the length of which (in kilometers) is given in the table:
Determine the length of the shortest path between points A and E. You can only move along the roads, the length of which is indicated in the table.
Explanation.
Find all route options from A to E and choose the shortest one.
From point A you can get to points B, C, D.
From point B you can go to point C.
From point C you can get to points D, E.
A-B-C-E: route length 7 km.
A-С-E: route length 7 km.
A-D-C-E: route length 6 km.
The correct answer is number 3.
10. A file was stored in some directory Lilac.doc, which had the full name D:\2013\Summer\Lilac.doc June and file Lilac.doc moved to the created subdirectory. Specify the full name of this file after moving.
1) D:\2013\Summer\Lilac.doc
2) D:\2013\Summer\June\Lilac.doc
Explanation.
The full filename after the move will be D:\2013\Summer\June\Lilac.doc.
11. A file was stored in some directory Lilac.doc. A subdirectory has been created in this directory June and file Lilac.doc moved to the created subdirectory. The full filename has become
D:\2013\Summer\June\Lilac.doc
Specify the full name of this file before moving.
1) D:\2013\Summer\Lilac.doc
2) D:\2013\Lilac.doc
3) D:\2013\Summer\June\Lilac.doc
Explanation.
The full name of the file before the move was D:\2013\Summer\Lilac.doc.
The correct answer is number 1.
12. Marina Ivanova, working on a project on literature, created the following files:
D:\Literature\Project\Yesenin.bmp
D:\Study\Work\Writers.doc
D:\Study\Work\Poets.doc
D:\Literature\Project\Pushkin. bmp
D:\Literature\Project\Poems.doc
Specify the full name of the folder, which will remain empty when deleting all files with the extension .doc. Assume that there are no other files and folders on drive D.
1) Literature
2) D:\Study\Work
3) D:\Study
4) D:\Literature\Project
Explanation.
Note that there are no other files in the Work folder other than Writers.doc And Poets.doc. Therefore, when deleting all files with the extension .doc, this folder will remain empty.
The correct answer is number 2.
Given a fragment of a spreadsheet:
From the diagram, you can see that the values in three cells are equal, and in the fourth one three times more. Since A2 = B2 ≠ D2, C2 = 3.
The found value of C2 corresponds to the formula indicated under the number 2.
14. Given a fragment of a spreadsheet:
It can be seen from the diagram that the values in three cells are equal, and the value in the fourth is three times greater than the sum of the values in the first three cells B2 = C2 = 1, therefore, D2 = 1.
The found value of D2 corresponds to the formula indicated under the number 2.
15. Given a fragment of a spreadsheet:
It can be seen from the diagram that the values in the three cells are equal. Since C2 = D2, therefore A2 = 3.
The found value A2 corresponds to the formula indicated under the number 4.
16. Performer The draftsman moves on the coordinate plane, leaving a trace in the form of a line. The draftsman can execute the command Move to ( a, b) (where a, b (x, y) to a point with coordinates (x + a, y + b). If numbers a, b positive, the value of the corresponding coordinate increases; if negative, decreases.
(4, 2)(2, −3) (6, −1).
Repeat k times
Team1 Team2 Team3
End
Team1 Team2 Team3 repeat k once.
Repeat 5 times
Move by (0, 1) Move by (−2, 3) Move by (4, −5) End
The coordinates of the point from which the Painter started moving, (3, 1). What are the coordinates of the point where he ended up?
Explanation.
Command Repeat 5 times means that the commands Move by (0, 1) Move by (−2, 3) Move by (4, −5) run five times. As a result, the Drafter will move 5 (0 − 2 + 4, 1 + 3 − 5) = (10, −5). Since the Drafter started moving at the point with coordinates (3, 1), the coordinates of the point where he ended up: (13, −4) .
The correct answer is number 3.
17. Performer The draftsman moves on the coordinate plane, leaving a trace in the form of a line. The draftsman can execute the command Move to ( a, b) (where a, b- integers) moving the Painter from the point with coordinates (x, y) to a point with coordinates (x + a, y + b). If numbers a, b positive, the value of the corresponding coordinate increases; if negative - decreases.
For example, if the Draftsman is at the point with coordinates (4, 2), then the command Move to(2, −3)will move the Draftsman to a point(6, −1).
Repeat k times
Team1 Team2 Team3
End
means that the sequence of commands Team1 Team2 Team3 repeat k once.
The draftsman was given the following algorithm to execute:
Repeat 3 times
End
What one command can be replaced by this algorithm so that the Draftsman is at the same point as after the execution of the algorithm?
1) Move by (−9, −3)
2) Move by (−3, 9)
3) Move by (−3, −1)
4) Move to (9, 3)
Explanation.
Command Repeat 3 times means that the commands Move by (−2, −3) Move by (3, 2) Move by (−4,0) will be executed three times. As a result, the Draftsman will move 3 (−2 + 3 − 4, −3 + 2 + 0) = (−9, −3). Thus, this algorithm can be replaced by the command Move by (−9, −3).
The correct answer is number 1.
18. Performer The draftsman moves on the coordinate plane, leaving a trace in the form of a line. The draftsman can execute the command Shift to (a, b) (where a, b are integers) moving the Painter from the point with coordinates ( x, y) to a point with coordinates ( x + a, y + b). If numbers a, b positive, the value of the corresponding coordinate increases; if negative, it decreases.
For example, if the Painter is at the point with coordinates (1, 1), then the Move by (-2, 4) command will move the Painter to the point (-1, 5).
Repeat k times
Team1 Team2 Team3
the end
means that the sequence of commands Team1 Team2 Team3 repeat k times.
The draftsman was given the following algorithm to execute:
Repeat 3 times
Move by (-2, -3) Move by (3, 4)
the end
Move by (-4, -2)
What command should the Drafter execute in order to return to the starting point from which he started moving?
1) Move by (1, -1)
2) Move by (-3, -1)
3) Move by (-3, -3)
4) Move by (-1, 1)
Explanation.
Command Repeat 3 times means that the commands Move by (-2, -3) and Move by (3, 4) will be executed three times. As a result, the Drafter will move 3 (−2 + 3, −3 + 4) = (3, 3). Thus, the draftsman will be at the point (3; 3), then he will execute the command Move by (-4, -2), after which it will end up at the point (−1; 1). Therefore, in order for the Drafter to return to the starting point, he needs to execute the command Move by (1, −1).
Answer: 1.
19. The following encrypted radiogram was received from the intelligence officer, transmitted using Morse code:
– – – – – – – –
During the transmission of the radiogram, the division into letters was lost, but it is known that only the following letters were used in the radiogram:
Some encryptions can be decrypted in more than one way. For example, 00101001 can mean not only URA, but also UAU. Three code chains are given:
Explanation.
1) "0100100101" can mean both "AUUA", and "PPAA", and "RAUA".
2) "011011111100" can only mean "ENTER".
3) "0100110001" can mean both "AUDA" and "RADA".
Answer: ENTER.
Answer: ENTER
21. Valya encrypts Russian words (sequences of letters), writing down its code instead of each letter:
| BUT | D | TO | H | ABOUT | FROM |
|---|---|---|---|---|---|
| 01 | 100 | 101 | 10 | 111 | 000 |
Some chains can be decrypted in more than one way. For example, 00010101 can mean not only SKA, but also SNK. Three code chains are given:
Find among them the one that has only one decryption, and write down the deciphered word in the answer.
Explanation.
Let's analyze each answer:
1) "10111101" can mean both "KOA" and "NOK".
2) "100111101" can mean both "DOK" and "HAOA".
3) "0000110" can only mean "SAN".
Hence the answer is "SAN".
Answer: SAN
22. In the program, ":=" denotes the assignment operator, the signs "+", "-", "*" and "/" - respectively, the operations of addition, subtraction, multiplication and division. The rules for performing operations and the order of actions correspond to the rules of arithmetic.
Determine the value of a variable b after executing the algorithm:
A:=8
b:= 3
a:= 3 * a – b
b:= (a / 3) * (b + 2)
In your answer, indicate one integer - the value of the variable b.
Explanation.
Let's execute the program:
A:=8
b:= 3
a:= 3 * 8 - 3 = 21
b:= (21 / 3) * (3 + 2) = 35
23. In the program, ":=" denotes the assignment operator, the signs "+", "-", "*" and "/" - respectively, the operations of addition, subtraction, multiplication and division. The rules for performing operations and the order of actions correspond to the rules of arithmetic. Determine the value of the variable b after executing the algorithm:
a:= 7
b:= 2
a:= b*4 + a*3
b:= 30 - a
Explanation.
Let's execute the program:
A:= 7
b:= 2
a:= b*4 + a*3 = 8 + 21 = 29
b:= 30 - a = 1.
24. The algorithm below uses the variables a and b. The symbol ":=" denotes the assignment operator, the signs "+", "-", "*" and "/" denote the operations of addition, subtraction, multiplication and division, respectively. The rules for performing operations and the order of actions correspond to the rules of arithmetic. Determine the value of the variable b after executing the algorithm:
a:= 5
b:= 2 + a
a:= a*b
b:= 2*a - b
In your answer, indicate one integer - the value of the variable b.
Explanation.
Let's execute the program:
A:= 5
b:= 2 + a = 7
a:= a*b=35
b:= 2*a - b = 63.
25. Determine what will be printed as a result of the following program. The text of the program is given in three programming languages.
Explanation.
The loop "for k:= 0 to 9 do" is executed ten times. Each time the variable s is increased by 3. Since initially s = 3, after executing the program we get: s = 3 + 10 3 = 33.
26. Determine what will be printed as a result of the following program. The text of the program is given in three programming languages.
Explanation.
The loop "for k:= 1 to 9 do" is executed nine times. Each time the variable s decreases by 3. Since initially s = 50, after the program is executed we get: s = 50 − 9 3 = 23.
27. Determine what will be printed as a result of the following program. The text of the program is given in three programming languages.
Explanation.
The "for k:= 1 to 7 do" loop is executed seven times. Each time the variable s is multiplied by 2. Since initially s = 1, after running the program we get: s = 1 2 2 2 2 2 2 2 = 128.
28. The table Dat presents data on the number of votes cast for 10 folk song performers (Dat - the number of votes cast for the first performer; Dat - for the second, etc.). Determine what number will be printed as a result of the following program. The text of the program is given in three programming languages.
| Algorithmic language | BASIC | Pascal |
|---|---|---|
|
alg |
DIM Dat(10) AS INTEGER |
Var k, m: integer; |
Explanation.
The program is designed to find the maximum number of votes cast for one performer. After analyzing the input data, we come to the conclusion that the answer is 41.
Answer: 41.
29. The Dat table stores data on the number of tasks completed by students (Dat tasks were done by the first student, Dat by the second, etc.). Determine what number will be printed as a result of the following program. The text of the program is given in three programming languages.
| Algorithmic language | BASIC | Pascal |
|---|---|---|
|
algnach m:= 10; n:=0 |
DIM Dat(10) AS INTEGER DIM k,m,n AS INTEGER IF Data(k)< m THEN m =Dat[k] |
Var k, m, n: integer; Dat: array of integers; m:= 10; n:=0; |
Explanation.
The program is designed to find the number of the student who made the least number of tasks. After analyzing the input data, we come to the conclusion that the answer is 4.
30. The Dat table stores marks of grade 9 students for independent work (Dat is the mark of the first student, Dat is the second, etc.). Determine what number will be printed as a result of the following program. The text of the program is given in three programming languages.
| Algorithmic language | BASIC | Pascal |
|---|---|---|
|
alg |
DIM Dat(10) AS INTEGER |
Var k, m: integer; |
Explanation.
The program is designed to find the sum of marks of students whose mark is less than four. After analyzing the input data, we come to the conclusion that the answer is the number 11.
Answer: 11.
31. In the figure - a diagram of the roads connecting the cities A, B, C, D, E, F, G, H. 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 H?
Explanation.
H can be reached from C, D, or G, so N = N H = N C + N D + N G (*).
Similarly:
N C \u003d N A + N D \u003d 1 + 3 \u003d 4;
N G = N D + N E + N F = 3 + 2 + 1 = 6;
N D \u003d N A + N E \u003d 1 + 2 \u003d 3;
N E \u003d N A + N B \u003d 1 + 1 \u003d 2;
Substitute in the formula (*): N = 4 + 3 + 6 = 13.
Answer: 13.
32. In the figure - a diagram of the roads connecting the cities A, B, C, D, D, E, K. You can move along each road only in one direction, indicated by the arrow. How many different routes are there from city A to city K?
Explanation.
Let's start counting the number of paths from the end of the route - from city K. Let N X be the number of different paths from city A to city X, N - the total number of paths.
You can come to K from E or D, so N = N K = N E + N D (*).
Similarly:
N D \u003d N B + N A \u003d 1 + 1 \u003d 2;
N E \u003d N B + N C + N G \u003d 1 + 2 + 3 \u003d 6;
N B \u003d N A \u003d 1;
N B \u003d N B + N A \u003d 1 + 1 \u003d 2;
N G \u003d N A + N B \u003d 1 + 2 \u003d 3.
Substitute in the formula (*): N = 2 + 6 = 8.
33. In the figure - a diagram of the roads connecting the cities A, B, C, D, E, F, G, H. 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 H?
Explanation.
Let's start counting the number of paths from the end of the route - from city H. Let N X be the number of different paths from city A to city X, N - the total number of paths.
H can be reached from E, F, or G, so N = N H = N E + N F + N G (*).
Similarly:
N E \u003d N A + N F \u003d 1 + 4 \u003d 5;
N G = N F + N D + N C = 4 + 3 + 1 = 8;
N F \u003d N A + N D \u003d 1 + 3 \u003d 4;
N D = N A + N B + N C = 1+ 1 + 1 = 3;
Substitute in the formula (*): N = 5 + 4 + 8 = 17.
Answer: 17.
34. Below in tabular form is a fragment of the database "Books of our store".
How many genres in this fragment satisfy the condition
(Number of books > 35) AND (Average cost< 300)?
In your answer, indicate one number - the desired number of genres.
Explanation.
The logical AND is true when both statements are true. Therefore, those options are suitable in which the number of books exceeds 35 and the average cost is less than 300 rubles. There are 2 such options.
Answer: 2.
35. Below in tabular form is a fragment of the database "Departure of long-distance trains":
| Destination | Train category | Travel time | Railway station |
|---|---|---|---|
| Baku | ambulance | 61:24 | Kursk |
| Balashov | passenger | 17:51 | Paveletsky |
| Balashov | passenger | 16:57 | Paveletsky |
| Balkhash | ambulance | 78:45 | Kazansky |
| Berlin | ambulance | 33:06 | Belorussian |
| Brest | ambulance | 14:47 | Belorussian |
| Brest | ambulance | 24:16 | Belorussian |
| Brest | accelerated | 17:53 | Belarusian |
| Brest | passenger | 15:45 | Belorussian |
| Brest | passenger | 15:45 | Belorussian |
| Valuyki | branded | 14:57 | Kursk |
| Varna | ambulance | 47:54 | Kievsky |
In your answer, indicate one number - the desired number of records.
Explanation.
The logical "OR" is true when at least one statement is true. Therefore, options are suitable in which the train is “passenger” and in which the station is “Belorussky”. There are 8 such options.
36. Below in tabular form is a fragment of the base on the tariffs of the Moscow metro.
How many records in this fragment satisfy the condition (Cost in rubles > 400) OR (Validity period< 30 дней)? In your answer, indicate one number - the desired number of records.
Explanation.
The logical "OR" is true when at least one statement is true. Therefore, options are suitable in which the fare is more than 400 rubles or the validity period is less than 30 days. There are 5 such options.
Answer: 5.
37. Convert the number 101010 from binary to decimal. Write down the resulting number in your answer.
Explanation.
Let's represent the number 101010 as a sum of powers of two:
101010 2 = 1 2 5 + 1 2 3 + 1 2 1 = 32 + 8 + 2 = 42.
38. Convert the number 68 from decimal to binary. How many units does the resulting number contain? In your answer, write one number - the number of units.
Explanation.
Let's represent the number 68 as a sum of powers of two: 68 = 64 + 4. Now we translate each of the terms into the binary system and add the results: 64 = 100 0000, 4 = 100. Therefore, 68 10 = 100 0100 2 .
Answer: 2.
39. Convert the binary number 1110001 to decimal.
Explanation.
1110001 2 = 1 2 6 + 1 2 5 + 1 2 4 + 1 2 0 = 64 + 32 + 16 + 1 = 113.
40. The performer Quadrator has two teams, which are assigned numbers:
1. add 3
2. square
The first of them increases the number on the screen by 3, the second raises it to the second power. The performer works only with natural numbers. Make an algorithm for obtaining from number 4 of number 58, containing no more than 5 commands. In the answer, write down only the numbers of the commands.
(For example, 22111 is the algorithm:
square
square
add 3
add 3
add 3,
which converts the number 3 to 90).
Explanation.
The closest number to 58 whose square root is an integer is 49 = 7 2 . Note that 58 \u003d 49 + 3 + 3 + 3. Let's sequentially go from the number 4 to the number 58:
4 + 3 = 7 (team 1);
7 2 = 49 (team 2);
49 + 3 = 52 (team 1);
52 + 3 = 55 (team 1);
55 + 3 = 58 (team 1).
Answer: 12111.
Answer: 12111
41. The performer Multiplier has two teams that are assigned numbers:
1. multiply by 3
2. subtract 1
The first of them multiplies the number by 3, the second - subtracts from the number 1. The performer works only with natural numbers. Make an algorithm for obtaining from number 8 of number 61, containing no more than 5 commands. In the answer, write down only the numbers of the commands.
(For example, 22112 is the algorithm:
subtract 1
subtract 1
multiply by 3
multiply by 3
subtract 1
which converts the number 5 to 26.
If there is more than one such algorithm, write down any of them.
Explanation.
Let's sequentially go from the number 8 to the number 61:
8 − 1 = 7 (team 2);
7 3 = 21 (team 1);
21 3 = 63 (team 1);
63 − 1 = 62 (team 2);
62 − 1 = 61 (team 2).
Answer: 21122.
Answer: 21122
42. The performer Multiplier has two teams that are assigned numbers:
1. multiply by 3
2. add 2
The first of them multiplies the number by 3, the second one adds 2 to the number. Make an algorithm for obtaining the number 58 from the number 2, containing no more than 5 commands. In the answer, write down only the numbers of the commands.
(For example, 21122 is the algorithm:
add 2
multiply by 3
multiply by 3
add 2
add 2,
which converts the number 1 to 31).
If there is more than one such algorithm, write down any of them.
Explanation.
Multiplication by a number is not reversible for any number, therefore, if we go from the number 58 to the number 2, then we will uniquely restore the program. Received commands will be written from right to left. If the number is not a multiple of 3, then subtract 2, and if it is a multiple, then divide by 3:
58 − 2 = 56 (team 2);
56 − 2 = 54 (team 2);
54 / 3 = 18 (team 1);
18 / 3 = 6 (team 1).
6 / 3 = 2 (team 1).
Let's write the sequence of commands in reverse order and get the answer: 11122.
Answer: 11122.
Answer: 11122
43. A 32 KB file is transmitted over some connection at a rate of 1024 bits per second. Determine the size of a file (in bytes) that can be transferred in the same amount of time over another connection at 128 bits per second. In your answer, enter a single number - the size of the file in bytes. Units of measurement are not required.
Explanation.
Transferred file size = transfer time · transfer rate. Note that the transmission rate in the second case is 1024/128 = 8 times less than the rate in the first case. Since the file transfer time is the same, the file size that can be transferred in the second case is also 8 times smaller. It will be equal to 32/8 = 4 KB = 4096 bytes.
Answer: 4096
44. A 2 MB file is transferred over some connection in 80 seconds. Determine the file size (in KB) that can be transferred over the same connection in 120 seconds. In your answer, indicate one number - the size of the file in Kbytes. Units of measurement are not required.
Explanation.
Transferred file size = transfer time · transfer rate. Note that the transmission time in the second case is 120/80 = 1.5 times the time in the first case. Since the file transfer speed is the same, the file size that can be transferred in the second case is also 1.5 times larger. It will be equal to 1.5 2048 = 3072 KB.
Answer: 3072
45. A 2000 KB file is transferred over some connection for 30 seconds. Determine the file size (in KB) that can be transferred over this connection in 12 seconds. In your answer, indicate one number - the size of the file in KB. Units of measurement are not required.
Explanation.
Let's calculate the data transfer rate over the channel: 2000 Kb / 30 sec = 200 / 3 Kb / sec. Therefore, the size of a file that can be transferred in 12 seconds is 200/3 KB/sec · 12 sec = 800 KB.
46. The machine receives a four-digit decimal number as input. Based on the received number, a new decimal number is constructed according to the following rules.
1. Two numbers are calculated - the sum of the first and second digits and the sum of the third and fourth digits of the given number.
2. The resulting two numbers are written one after the other in non-decreasing order (without separators).
Example. Original number: 2177. Bitwise sums: 3, 14. Result: 314.
Determine how many of the numbers below can be obtained as a result of the operation of the machine.
1915 20 101 1213 1312 312 1519 112 1212
Write down only the number of numbers in your answer.
Explanation.
Let's analyze each number.
The number 1915 cannot be the result of the machine, since the number 19 cannot be obtained by adding two digits.
The number 20 cannot be the result of the automaton, since the resulting two numbers are written one after the other in non-decreasing order.
The number 101 cannot be the result of the automaton, since its first part is 1, and the second part is 01, which is not a number.
The number 1213 may be the result of the machine, in which case the original number could be 6667.
The number 1312 cannot be the result of the machine's operation, since the two numbers obtained are written one after the other in non-decreasing order.
The number 312 may be the result of the machine, in which case the original number could be 2166.
The number 1519 cannot be the result of the machine, since the numbers are written in non-decreasing order, and the number 19 cannot be obtained by adding two digits.
The number 112 may be the result of the machine, in which case the original number could be 1057.
The number 1212 may be the result of the machine, in which case the original number could be 6666.
47. A chain of four beads marked in Latin letters is formed according to the following rule:
- in the third place of the chain is one of the beads H, E;
- in the second place - one of the beads D, E, C, which is not in the third place;
- at the beginning there is one of the beads D, H, B, which is not in the second place;
- at the end - one of the beads D, E, C, which is not in the first place.
Determine how many of the listed chains are created according to this rule?
DEHD HEHC DCEE DDHE DCHE HDHD BHED EDHC DEHE
In your answer, write down only the number of chains.
Explanation.
First chain DEHD does not satisfy the fourth condition of the rule, the fourth DDHE- to the third. Seventh chain BHED does not satisfy the second condition of the rule. Eighth chain EDHC does not satisfy the third condition of the rule.
Thus, we have five chains that satisfy the condition.
48. Some algorithm from one string of characters gets a new string as follows. First, the length of the original character string is calculated; if it is even, then the last character of the chain is deleted, and if it is odd, then the character C is added to the beginning of the chain. In the resulting chain of characters, each letter is replaced by the letter following it in the Russian alphabet (A - to B, B - to C, etc.). D., and I - on A). The resulting chain is the result of the algorithm.
For example, if the original string was LEG OPD, and if the original chain was TONE, then the result of the algorithm will be the chain STUPIDLY.
Given a string of characters RAFT. What string of characters will be obtained if the described algorithm is applied to this string twice (i.e., apply the algorithm to this string, and then apply the algorithm again to the result)? Russian alphabet: ABVGDEYOZHZIYKLMNOPRSTUFKHTSCHSHCHYYYYYUYA.
Explanation.
Let's apply the algorithm: RAFT(even) → PLO → RMP.
Let's apply it again: RMP(odd) → SRMP → TSNR.
Answer: TSNR
49. File Access com.txt mail.nethttp
Explanation.
http://mail.net/com.txt. Hence, the answer is BVEDAGG.
Answer: BVEDAGG
50. File Access doc.htm located on the server site.com, carried out according to the protocol http. Fragments of the file address are encoded with letters from A to G. Write down the sequence of these letters that encodes the address of the specified file on the Internet.
Explanation.
Recall how an address is formed on the Internet. First, the protocol is indicated (usually it is “ftp” or “http”), then “://”, then the server, then “/”, the file name is indicated at the end. So the address will be: http://site.com/doc.htm. Therefore, the answer is ZHBAEGVD.
Answer: ZHBAEGVD
51. File Access rus.doc located on the server obr.org, carried out according to the protocol https. Fragments of the file address are encoded with letters from A to Z. Write down the sequence of these letters that encodes the address of the specified file on the Internet.
Explanation.
Recall how an address is formed on the Internet. First, the protocol is indicated (usually it is “ftp” or “http”), then “://”, then the server, then “/”, the file name is indicated at the end. So the address will be: https://obr.org/rus.doc. Hence, the answer is JGAVBED.
Answer: ZGAVBED
52. The table shows queries to the search server. Arrange the query designations in ascending order of the number of pages that the search engine will find for each query. To indicate the logical operation "OR" in the query, the symbol "|" is used, and for the logical operation "AND" - "&":
Explanation.
The more "OR" in the query, the more results the search server produces. The more "AND" operations in the query, the fewer results the search server will give. So the answer is BVAG.
Answer: BVAG
53. The table shows queries to the search server. For each request, its code is indicated - the corresponding letter from A to D. Arrange the request codes from left to right in ascending order of the number of pages that the search engine found for each request. For all queries, a different number of pages were found. To indicate the logical operation "OR" in the query, the symbol "|" is used, and for the logical operation "AND" - "&":
Explanation.
The more "OR" in the query, the more results the search server produces. The more "AND" operations in the query, the fewer results the search server will give. So the answer is GBVA.
Answer: GBVA
54. The table shows queries to the search server. Arrange the query designations in ascending order of the number of pages that the search engine will find for each query. To indicate the logical operation "OR" in the query, the symbol "|" is used, and for the logical operation "AND" - "&":
Explanation.
The more "OR" in the query, the more results the search server produces. The more "AND" operations in the query, the fewer results the search server will give. Thus, the answer is AGBV.
Answer: AGBV
55. The results of passing the standards in athletics among students in grades 7-11 were entered into the spreadsheet. The figure shows the first lines of the resulting table:
Column A contains the last name; in column B - name; in column C - gender; in column D - year of birth; in column E - the results in the run for 1000 meters; in column F - the results in the 30 meters run; in column G - standing long jump results. In total, data on 1000 students were entered into the spreadsheet.
Complete the task.
1. What percentage of the participants showed results in long jumps over 2 meters? Write your answer in cell L1 of the table.
2. Find the difference in seconds, to the nearest tenth, between the average of the participants born in 1996 and the average of the participants born in 1999 in the 30 meters. Write the answer to this question in cell L2 of the table.
Complete the task.
Open the spreadsheet file. Based on the data in this table, answer two questions.
1. How many days during this period was atmospheric pressure above 760 mmHg? Write the answer to this question in cell H2 of the table.
2. What was the average wind speed on days with air temperatures below 0 °C? Write the answer to this question with an accuracy of at least 2 decimal places in cell H3 of the table.
Explanation.
Solution for OpenOffice.org Calc and for Microsoft Excel
The first formula is used for the Russian-language notation of functions, the second - for the English-language one.
In cell H2, we write a formula that determines how many days during a given period the atmospheric pressure was above 760 mm Hg:
COUNTIF(C2:C397;">760")
=COUNTIF(C2:C397;">760")
To answer the second question in the cell, in column G for each day, we write the wind speed if the air temperature is below 0 °C on that day, and "" otherwise. In cell G2, write the formula
IF(B2<0;D2; «»)
=IF(B2<0;D2; «»)
Copy the formula to all cells in the G2:G397 range. Next, to determine the average wind speed, we write the formula in cell H3:
AVERAGE(G2:G397)
=AVERAGE(G2:G397)
Other ways of solving the problem are also possible.
If the task was completed correctly and when the task was executed, files specially prepared for checking the execution of this task were used, then the following answers should be obtained:
to the first question: 6;
to the second question: 1.67.
57. Data on testing students was entered into the spreadsheet. Below are the first five rows of the table:
Column A lists the district in which the student is studying; in column B - last name; in column C - favorite subject; in column D is the test score. In total, data on 1000 students were entered into the spreadsheet.
Complete the task.
Open the file with this spreadsheet (the location of the file will be communicated to you by the exam organizers). Based on the data in this table, answer two questions.
1. How many students in District Northeast (NE) chose math as their favorite subject? Write the answer to this question in cell H2 of the table.
2. What is the average test score for students in the Southern District (S)? Write the answer to this question in cell H3 of the table with an accuracy of at least two decimal places.
Explanation. task19.xls
1. Write in cell H2 the following formula =IF(A2="CB";C2;0) and copy it to the range H3:H1001. In this case, the name of the subject will be written in the cell of the H column if the student is from the North-Eastern district and "0" if it is not. By applying the operation =IF(H2="Math",1,0), we get a column (J) with ones and zeros. Next, we use the operation =SUM(J2:J1001). Let's get the number of students who consider mathematics their favorite subject. There are 17 such students.
2. To answer the second question, use the "IF" operation. Let's write the following expression in cell E2: =IF(A2="Yu",D2,0), as a result of applying this operation to the range of cells E2:E1001, we get a column in which the scores of only the students of the Southern District are recorded. Summing up the values in the cells, we get the sum of the students' scores: 66,238. Next, we calculate the number of students in the Southern District using the command =COUNTIF(A2:A1001, "YU"), we get: 126. Dividing the sum of points by the number of students, we get: 525.69 - the required average score.
Answer: 1) 17; 2) 525.70.
20.1
The Robot has nine commands. Four commands are commands-orders:
up down left right
When any of these commands is executed, the Robot moves one cell respectively: up , down ↓, left ←, right →. If the Robot receives a command to move through the wall, it will collapse. The robot also has a team paint over
Four more commands are commands for checking conditions. These commands check if the path is clear for the Robot in each of the four possible directions:
top free bottom free left free right free
These commands can be used in conjunction with the " if", having the following form:
if condition then
command sequence
all
Here condition– one of the condition check commands.
Command sequence- this is one or more of any commands-orders.
For example, to move one cell to the right, if there is no wall on the right and painting the cell, you can use the following algorithm:
if the right is free then
right
paint over
all
In one condition, you can use several commands for checking conditions, using logical connectives and, or, not, for example:
right
all
« till", which looks like this:
nts bye condition
command sequence
kts
nc while right free
right
kts
Complete the task.
There is a wall on the endless field. The wall consists of three consecutive segments: to the right, down, to the right, all segments of unknown length. The robot is in the cage located immediately above the left end
first cut. The figure shows one of the possible ways to arrange the walls and the Robot (the Robot is indicated by the letter "P").
Write an algorithm for the Robot that fills all the cells immediately to the right of the second segment and above the third one. The robot must paint over only the cells that meet this condition. For example, for the above picture, the Robot must paint over the following cells (see picture).
The final location of the Robot can be arbitrary. The algorithm must solve the problem for an arbitrary field size and any admissible location of walls inside a rectangular field. When executing the algorithm, the Robot should not collapse.
20.2 Write a program that, in a sequence of natural numbers, finds the arithmetic mean of multiples of 8, or reports that there are no such numbers (outputs "NO"). The program receives natural numbers as input, the number of entered numbers is unknown, the sequence of numbers ends with the number 0 (0 is a sign of the end of the input, is not included in the sequence).
The number of numbers does not exceed 100. The entered numbers do not exceed 300. The program should display the arithmetic mean of multiples of 8, or print "NO" if there are no such numbers. Display the value with an accuracy of tenths.
An example of the program's operation:
| Input data | Output |
| 8 122 64 16 0 |
29,3 |
| 111 1 0 |
NO |
Explanation.
20.1 The executor's commands will be written in bold, and the comments explaining the algorithm and not being part of it will be in italics. The beginning of the comment will be denoted by the symbol "|".
| Move right along the top horizontal wall until it ends.
nc not yet (bottom free)
right
kts
| Move down along the vertical wall and paint the cells
nc while the bottom is free
down
paint over
kts
| Move to the right along the horizontal wall and paint the cells
nc not yet (bottom free)
paint over
right
kts
20.2 A solution is a program written in any programming language. An example of a correct solution written in Pascal:
var a, s, n: integer;
begin
s:=0; n:=0;
readln(a);
while a<>0 to begin
if (a mod 8 = 0) then
begin
s:= s + a;
n:= n + 1;
end;
readln(a); end;
if n > 0 then writeln(s/n:5:1)
else writeln('NO');
end.
Other solutions are also possible. To check the correct operation of the program, you must use
the following tests:
| № | Input data | Output |
| 1 | 2 222 0 |
NO |
| 2 | 16 0 |
16.0 |
| 3 | 1632 64 8 8 5 0 |
25.6 |
59. Choose ONE of the tasks below: 20.1 or 20.2.
20.1 The Executor Robot is able to move through the labyrinth drawn on a plane divided into cells. Between adjacent (on the sides) cells there can be a wall through which the Robot cannot pass.
The Robot has nine commands. The four commands are command commands:
up down left right
When any of these commands is executed, the Robot moves one cell respectively: up down ↓, left ← , right →. If the Robot receives a command to move through the wall, it will collapse.
The robot also has a team paint over, at which the cell in which the Robot is currently located is painted over.
Four more commands are commands for checking conditions. These commands check if the path is clear for the Robot in each of the four possible directions:
These commands can be used together with the condition "if", having the following form:
if condition then
command sequence
all
Here condition- one of the condition checking commands. Command sequence- this is one or more of any commands-orders. For example, to move one cell to the right, if there is no wall on the right, and paint the cell, you can use the following algorithm:
if the right is free then
right
paint over
all
In one condition, you can use several commands for checking conditions, using logical connectives and, or, not, for example:
if (right free) and (not bottom free) then
right
all
You can use a loop to repeat a sequence of commands. "till", which looks like this:
nts bye condition
command sequence
kts
For example, to move to the right while possible, you can use the following algorithm:
nc while right free
right
kts
Complete the task.
The infinite field has horizontal and vertical walls. The left end of the horizontal wall is connected to the bottom end of the vertical wall. The lengths of the walls are unknown. There is exactly one passage in the vertical wall, the exact location of the passage and its width are unknown. The robot is in a cage located directly above the horizontal wall at its right end. The figure shows one of the possible ways to arrange the walls and the Robot (the Robot is indicated by the letter "P").
Write an algorithm for the Robot that paints all the cells immediately to the left and to the right of the vertical wall.
The robot must paint over only the cells that meet this condition. For example, for the picture on the right, the Robot must paint over the following cells (see picture).
The final location of the Robot can be arbitrary. When executing the algorithm, the Robot should not collapse. The algorithm must solve the problem for an arbitrary field size and any admissible wall arrangement.
The algorithm can be executed in the environment of a formal executor or written in a text editor.
20.2
Write a program that, in a sequence of natural numbers, determines the minimum number ending in 4. The program receives as input the number of numbers in the sequence, and then the numbers themselves. The sequence always contains a number ending in 4. The number of numbers does not exceed 1000. The entered numbers do not exceed 30,000. The program should print one number - the minimum number,
ending in 4.
An example of the program's operation:
| Input data | Output |
| 14 |
Explanation.20.1 The executor's commands will be written in bold, and the comments explaining the algorithm and not being part of it will be in italics. The beginning of the comment will be denoted by the symbol "|".
||We move to the left until we reach a vertical wall.
nc while left free
to the left
kts
|We move up until we reach the passage in the wall, and paint over the cells.
nc until left loose
paint over
up
kts
nc while left free
up
kts
|Move up to the end of the wall and paint over the cells.
nc until left loose
paint over
up
kts
| Go around the wall. |
to the left
down
|We move down until we reach the passage in the wall, and paint over the cells.
nc until right loose
paint over
down
kts
|We move on to the vertical wall.
nc while right free
down
kts
|We move down to the end of the wall and paint over the cells.
nc until right loose
paint over
down
kts
Other solutions are also possible. It is allowed to use a different syntax for the instructions of the executor,
more familiar to students. It is allowed to have separate syntax errors that do not distort the intention of the author of the decision
20.2 The solution is a program written in any programming language. An example of a correct solution written in Pascal:
Varn,i,a,min: integer;
begin
readln(n);
min:= 30001;
for i:= 1 to n do
begin
readln(a);
if (a mod 10 = 4) and (a< min)
then min:= a;
end;
writeln(min)
end.
Other solutions are also possible. To check the correct operation of the program, you must use the following tests:
| № | Input data | Output |
|---|---|---|
| 1 | 4 | |
| 2 | 14 | |
| 3 | 4 |
60. Choose ONE of the tasks below: 20.1 or 20.2.
20.1 The Executor Robot is able to move through the labyrinth drawn on a plane divided into cells. Between adjacent (on the sides) cells there can be a wall through which the Robot cannot pass. The Robot has nine commands. The four commands are command commands:
up down left right
When any of these commands is executed, the Robot moves one cell respectively: up down ↓, left ← , right →. If the Robot receives a command to move through the wall, it will collapse. The robot also has a team paint over, at which the cell in which the Robot is currently located is painted over.
Four more commands are commands for checking conditions. These commands check if the path is clear for the Robot in each of the four possible directions:
top free bottom free left free right free
These commands can be used together with the condition "if", having the following form:
if condition then
command sequence
all
Here condition- one of the condition checking commands. Command sequence- this is one or more of any commands-orders. For example, to move one cell to the right, if there is no wall on the right, and paint the cell, you can use the following algorithm:
if the right is free then
right
paint over
all
In one condition, you can use several commands for checking conditions, using logical connectives and, or, not, for example:
if (right free) and (not bottom free) then
right
all
You can use a loop to repeat a sequence of commands. "till", which looks like this:
nts bye condition
command sequence
kts
For example, to move to the right while possible, you can use the following algorithm:
nc while right free
right
kts
Complete the task.
There is a staircase on the endless field. First, the ladder goes up from left to right, then goes down also from left to right. To the right of the descent, the staircase passes into a horizontal wall. The height of each step is 1 cell, the width is 1 cell. The number of steps leading up and the number of steps leading down is unknown. Between the descent and the ascent, the width of the site is 1 cell. The robot is in a cage located at the beginning of the descent. The figure shows one of the possible ways of arranging the walls and the Robot (the Robot is indicated by the letter "P").
Write an algorithm for the Robot that paints all the cells directly above the stairs. The robot must paint over only the cells that meet this condition. For example, for the above picture, the Robot must paint over the following cells (see picture).
The final location of the Robot can be arbitrary. The algorithm must solve the problem for an arbitrary field size and any admissible location of walls inside a rectangular field. When executing the algorithm, the Robot should not be destroyed, the execution of the algorithm should be completed. The algorithm can be executed in the environment of a formal executor or written in a text editor. Save the algorithm in a text file.
20.2 Enter 8 positive integers from the keyboard. Determine how many of them are divisible by 3 and end in 4. The program should output one number: the number of numbers that are multiples of 3 and end in 4.
An example of the program's operation:
| Input data | Output |
| 12 14 24 54 44 33 84 114 |
4 |
Explanation.20.1 The following algorithm will perform the required task.
nc until right loose
paint over
up
paint over
right
kts
paint over
right
nc while the bottom is free
paint over
down
paint over
right
kts
20.2 Solution
Var i, n, a: integer;
beginn:=0;
for i := 1 to 8 do
begin
readln(a);
if (a mod 3 = 0) and (a mod 10 = 4) then
n: = n + 1; end;
writeln(n);
end.
To check the correct operation of the program, you must use the following tests:
| Input data | Output | |
|---|---|---|
| 1 | 0 | |
| 2 | 1 | |
| 3 | 3 |
OGE tasks in computer science with solutions and answers
The 2019 state final certification in informatics for graduates of the 9th grade of general educational institutions is carried out in order to assess the level of general education of graduates in this discipline. The main elements of content from the informatics section that are checked in testing:
- The ability to evaluate the quantitative parameters of information objects.
- The ability to determine the value of a logical expression.
- Ability to analyze formal descriptions of real objects and processes.
- Knowledge of file system data organization.
- Ability to represent formula dependence in graphical form.
- The ability to execute an algorithm for a specific performer with a fixed set of commands.
- Ability to encode and decode information.
- The ability to execute a linear algorithm written in an algorithmic language.
- The ability to execute the simplest cyclic algorithm written in an algorithmic language.
- The ability to execute a cyclic algorithm for processing an array of numbers, written in an algorithmic language.
- Ability to analyze information presented in the form of diagrams.
- Ability to search in a ready-made database according to the formulated condition.
- Knowledge of the discrete form of representation of numerical, textual, graphic and sound information.
- Ability to write a simple linear algorithm for a formal performer.
- The ability to determine the speed of information transfer.
- The ability to execute an algorithm written in natural language that processes strings of characters or lists.
- Ability to use information and communication technologies.
- Ability to search for information on the Internet.
- Ability to process large amounts of data using spreadsheet or database tools.
- Ability to write a short algorithm in the environment of a formal executor or in a programming language.
| Dates for passing the OGE in Informatics 2019: June 4 (Tuesday), June 11 (Tuesday). |
There are no changes in the structure and content of the examination paper in 2019 compared to 2018. |
The standard OGE test (GIA-9) of the 2019 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMM), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2019 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMM), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2018 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMM), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2018 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2018 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2018 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2017 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2016 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2016 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2016 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2016 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2015 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2015 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
The standard OGE test (GIA-9) of the 2015 format in informatics and ICT contains two parts. The first part contains 18 tasks with a short answer, the second part contains 2 tasks that must be completed on the computer. In this regard, only the first part (the first 18 tasks) is presented in this test. According to the current structure of the exam, among these 18 tasks, answers are offered only in the first 6 tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each task. However, for tasks in which answer options are not provided by the compilers of real control and measuring materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will have to face at the end of the school year.
For tasks 1-18, choose only one correct answer.
For tasks 1-8, choose only one correct answer.
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Option 1
19. 1)38% 2)55
20. Task C2 No. 100
Assessment Criteria for Assignment 20.1 | Points |
The algorithm works correctly for all valid input data | |
For all admissible initial data, the following is true: 1) the algorithm execution is completed, and the Robot does not crash; 2) no more than 10 extra cells are shaded; 3) no more than 10 cells remained unpainted from among those that should have been painted over | |
The task was completed incorrectly, i.e., the conditions that allow you to put 1 or 2 points are not met | |
Maximum score |
Criteria for assessing the performance of task 20.2 | Points |
Proposed correct solution. The program works correctly on all the above tests. The program can be written in any programming language | |
The program gives an incorrect answer on one of the tests above | |
The program gives incorrect answers on tests that differ from those described in the criteria by 1 point | |
Maximum score |
The Executor Robot is able to move through the labyrinth drawn on a plane divided into cells. Between adjacent (on the sides) cells there can be a wall through which the Robot cannot pass. The Robot has nine commands. The four commands are command commands:
up down left right
When any of these commands is executed, the Robot moves one cell, respectively: up down ↓ , left ← , right → . If the Robot receives a command to move through the wall, it will collapse. The robot also has a team paint over , at which the cell in which the Robot is currently located is painted over.
Four more commands are commands for checking conditions. These commands check if the path is clear for the Robot in each of the four possible directions:
top free bottom free left free right free
These commands can be used together with the condition"if" , having the following form:
if condition then
command sequence
all
Here the condition - one of the condition checking commands.Command sequence- this is one or more of any commands-orders. For example, to move one cell to the right, if there is no wall on the right, and paint the cell, you can use the following algorithm:
if the right is free then
right
paint over
all
In one condition, you can use several commands for checking conditions, using logical connectives and, or, not, for example:
if (right free) and (not bottom free) then
right
all
You can use a loop to repeat a sequence of commands."till" , which looks like this:
nc bye condition
command sequence
kts
For example, to move to the right while possible, you can use the following algorithm:
nc while right free
right
kts
Complete the task.
The infinite field has horizontal and vertical walls. The right end of the horizontal wall is connected to the bottom end of the vertical wall. The lengths of the walls are unknown. There is exactly one passage in each wall, the exact location of the passage and its width are unknown. The robot is in a cage located immediately to the right of the vertical wall at its upper end. The figure shows one of the possible ways to arrange the walls and the Robot (the Robot is indicated by the letter "P").
Write an algorithm for the Robot that paints all the cells directly above the horizontal wall and to the left of the vertical wall. Passages must remain unpainted. The robot must paint over only the cells that meet this condition. For example, for the above picture, the Robot must paint over the following cells (see picture).
When executing the algorithm, the Robot should not be destroyed, the execution of the algorithm should be completed. The final location of the Robot can be arbitrary. The algorithm must solve the problem for any admissible wall arrangement and any location and size of passages inside the walls. The algorithm can be executed in the environment of a formal executor or written in a text editor. Save the algorithm in a text file.
20.2 Write a program that, in a sequence of natural numbers, determines the sum of numbers that are multiples of 3. The program receives as input the number of numbers in the sequence, and then the numbers themselves. The sequence always contains a number that is a multiple of 3. The number of numbers does not exceed 100. The entered numbers do not exceed 300. The program should output one number - the sum of numbers that are multiples of 3.
An example of the program's operation:
Input data | Output |
3 |
Explanation.
The following algorithm will perform the required task.
nc
until the left is free
paint over
down
Demo version of the OGE in Informatics 2018 + answers and criteria
Characteristics of the structure and content of KIM OGE 2018 in informatics
The OGE in Informatics and Information and Communication Technologies consists of 2 parts: written and practical (computing tasks on a computer).
The number of workstations equipped with a computer should correspond to the number of exam participants in the audience.
Part 2 of the CMM is performed on a computer. The verifiable output of the Part 2 task is a file.
The tasks of this part involve the practical work of students at the computer using special software. The result of the execution of each task is a separate file.
Programs familiar to students should be installed on the computer.
The task of the 2nd part is given in two versions at the choice of the student:
The first version of the task provides for the development of an algorithm for the "Robot" executor (it is recommended to use the "Robot" executor's learning environment. For example, the "Kumir" learning development environment developed at NIISI RAS can be used as such an environment (http://www.niisi. ru/kumir), or any other environment that allows modeling the "Robot" executor. If the syntax of the executor's commands in the environment used differs from that given in the task, it is allowed to make changes to the text of the task in the part of the description of the "Robot" executor. in the absence of a learning environment for the "Robot" performer, the task solution is written in a simple text editor);
The second version of the task provides for writing the algorithm in the programming language being studied (if the study of the topic "Algorithmization" is carried out using a programming language). In this case, the programming system used in training is required to complete the task.
The execution of each task of part 2 is a separate file prepared in the appropriate program (text editor or spreadsheet). Exam participants save these files in a directory under the names specified by the exam organizers (technician).
In the Answer Forms (after completing the work on the computer), the names of the files with completed tasks are entered, including a unique number (KIM number).
The maximum number of points that an examinee can receive for completing the entire examination paper is 22 points.
The results of the OGE exam in computer science in the 9th grade can be used when enrolling students in specialized classes of a secondary school. The benchmark for selection in specialized classes can be an indicator, the lower limit of which corresponds to 15 points.
