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Topic: Solving a task of an increased level of complexity C1 (3 points)

A task. It was necessary to write a program that reads the coordinates of a point on the plane from the keyboard (x, y are real numbers) and determines whether this point belongs to a given shaded area (including boundaries). The area is limited by the hyperbola xy = 1 and the straight lines x = 2 and y = 2. The programmer was in a hurry and wrote the program incorrectly. varx,y: real; begin readln(x, y); if x*y>=1 then if x

Characteristics of the task of an increased level of complexity C1 Task C1 is aimed at testing the ability to read a short simple program in an algorithmic language (programming language) and the ability to search for and correct errors in a small program fragment. You are offered a certain task and a variant of its solution, in which (consciously) a mistake was made. You need to find this error and indicate how you need to change the program so that it correctly solves the problem. Please note that in the task you need to find a semantic, not a syntax error.

Consider examples In this problem, the shaded area is limited by three straight lines: horizontal (ox axis), vertical (oy axis), oblique (y=1/2 x-1) We get the condition: (x>=0) and (y=0.5 * x-1)

(y=0) a) b) (y>=x*x-2) and (y

x *x+y*y=0) (x *x+y*y=0) and (y=4) and (x=0) and (y

g) h) i) j) (x *x+y*y>=4) and (x>=-2) and (y>=0) and (y=4) and (x>=-2) and (x=0) and (y=1) and (x>=-2) and (x =-2) and (y =-2) and (x 0) and (y

(x *x+y*y=x) or (x

a) b) c) d) (x *x+y*y=-x) or (y=x) or (y>=0)) (y>=x*x-2) and ((y=0 ))

e) (x *x+y*y=0) or (y>=0))

Periodic functions (y>=0) and (y=0) and (x=-pi/2) and (y>=-1) and (y>=x-1) and (y

c) d) (y>=0) and (y=0) and (x=1.57)) (y>=0) and (y=0) and (x

Task 1. It was required to write a program, during which the coordinates of a point on the plane are read from the keyboard (x, y are real numbers) and the belonging of this point to a given shaded area (including borders) is determined. The area is limited by the hyperbola xy = 1 and the straight lines x = 2 and y = 2. The programmer was in a hurry and wrote the program incorrectly. varx,y: real; begin readln(x, y); if x*y>=1 then if x

Solution: x=0, y=0 (Any pair (x,y) that satisfies: xy 2 or (xy≥1 and x =1) and (x>0) and (x0) and (y

Problem 2. var x,y: real; begin readln(x, y); if x*y=-2 then write("belongs") else write("does not belong") end. Do the following in sequence: 1) Give an example of such numbers x, y, for which the program incorrectly solves the problem. 2) Indicate how the program needs to be improved so that there are no cases of its incorrect operation. (This can be done in several ways, just specify any way to refine the original program).

Solution: x=0, y=0 (Any pair (x,y) that satisfies: xy>-1 or x>2 or (xy≤-1 and x 0) 2) Possible modification: var x,y: real; begin readln(x, y); if (x*y0) and (x= -2) and (y

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Today we will analyze the theoretical aspect of 1 task from USE in Informatics. In this task, you need to be able to translate numbers from different number systems to others. The main number systems are: binary, octal, decimal (our native) and hexadecimal.

Converting numbers from the binary system to the hexadecimal number system.

First you need to write the following table in your draft:

Let's look at this table. The first column contains numbers from 0 to 15 in our native decimal number system. In the second column there are also numbers from 0 to 15, but in the binary system, and in the third also from 0 to 15 in the hexadecimal number system.

Write numbers from 0 to 15 in our native decimal system, no one takes out any difficulties.

Numbers in the binary system are best written according to the following rule: in the least significant bit we alternate zero and one, in the next bit the alternation of zeros and ones is twice as slow (two zeros, two ones, two zeros, etc.), in the next bit the alternation is even twice as slow (4 zeros, 4 units, etc.) and finally 8 zeros and 8 ones - in the most significant bit.

In the hexadecimal number system, in addition to our usual characters from 0 to 9, invented symbols A,B,C,D,E,F, and from these 16 characters (from 0 to 15) any number is made up, just as in our system any number of ten digits (from 0 to 9) is made up. Accordingly, in order to count from 0 to 15, you need to go through all the characters that available in hexadecimal (0 to F).

Now let's look at how to use this table to translate from binary to hexadecimal. Let's translate the number 100101000 from binary to hexadecimal.

To complete this task, we need to split our binary number into 4 digits starting from the right edge, and each 4 digits must be found in our table: 1000 - this will be 8, 0010 - 2, 0001 is 1. In the high order we have there is one unit left, we supplemented it with 3 zeros.

So the number 100101000 2 in binary would be 128 16 in hexadecimal.

Converting numbers from binary to octal
number system.

From binary to octal X 2 -> X 8 we translate in the same way, only now we take from the table not four digits, but three numbers.

So the number 1001111001 2 in the binary system will be equal to 1171 8 in the octal system.

Converting numbers from hexadecimal to binary
number system.

We do exactly the same as when converting numbers from binary to hexadecimal, but in reverse order. We look at the table: D - 1101, F - 1111, 4 - 0100. It turns out the number 010011111101. On the left we discard zeros 10011111101 .

4FD 16 -> 10011111101 2 .

Converting numbers from octal to binary
number system.

We proceed as we did before. We break each digit of the octal system into 3 digits of the binary system, using the table that is given at the beginning of the article. We discard zeros on the left.

347 8 -> 11100111 2 .

Converting numbers from binary to decimal
number system.

Let's translate the number:

We take the digits of a binary number, starting with the least significant digit (i.e., on the right), and begin to multiply by two to the appropriate degree. The degree starts from zero and increases by 1 each time. All these products are summable.

After the calculation, we get the number in the decimal system:

Result 11010011 2 -> 211 10

Converting numbers from decimal to binary
number system.

Consider how to convert from decimal to binary. Let's take a number 213 .

Converting numbers from hexadecimal to octal
reckoning and vice versa.

Let's translate the number A10 from hexadecimal to octal A10 16 -> X 8.

We break each digit of the hexadecimal code into 4 digits of the binary code from the table at the beginning of the article (i.e. we translate the number into binary). We divide the resulting number into three digits - and collect the number already in the octal system - as shown in the figure. We translate it back in the same way, only in reverse order.

Converting numbers from hexadecimal to decimal
number system.

Let's translate the number 5B3 from hexadecimal to decimal 5B3 16 -> X 10.

We act in the same way as when converting from binary to decimal, only multiply the numbers by 16 to the appropriate degree. We turn the letters into decimal numbers from the table. We start, as always, on the right, i.e. from the junior level.

Converting numbers from decimal to hexadecimal
number system.

Let's translate the number 203 from decimal to hexadecimal 203 10 -> X 16

We divide the number by 16 until we get a number from 1 to 15. We write the remainders in reverse order. We turn numbers from 10 to 15 into numbers.

Converting numbers from octal to decimal
number system.

Let's translate the number 347 from octal to decimal 347 8 -> X 10

We do the same as the previous examples, only now we multiply by 8 to the appropriate degree.

Converting numbers from decimal to octal
number system.

We do the same as the previous examples.

Happy exams!

The lesson considered solution 1 task of the Unified State Examination in Informatics 2017: a detailed explanation and analysis of tasks is given

The 1st topic is characterized as tasks basic level complexity, execution time - about 1 minute, maximum score — 1

Number systems and representation of information in PC memory

To solve task 1, you should remember and repeat the following topics:

Binary number system

Number of digits or system base: 2
Numbers (alphabet): 0, 1

Translation of numbers from the 10th system. c-i to binary

Translation of numbers from the 2nd system. count in the 10th

When working with large numbers, it is better to use:

Expansion in powers of two

Octal number system

8
Numbers (alphabet): 0, 1, 2, 3, 4, 5, 6, 7

Translation of numbers from the 10th system. count in the 8th

Translation of numbers from the 8th system. count in the 10th

Translation of numbers from the 8th system. mid to 2nd and back with triads

Translation from the 8th syst. mid to 2nd and back with triads

Hexadecimal number system

Number of digits or system base: 16
Numbers (alphabet): 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A (10), B (11), C (12), D (13), E (14), F (15)

Translation from the 10th syst. count in the 16th

Translation from the 16th number system to the 10th

Translation from the 16th syst. count in the 10th

Translation of numbers from the 2nd system. sch-i to the 16th and back in tetrads

Translation from 2nd p. sch-i to the 16th and back in tetrads

Usefulness for the binary number system:

• numbers that in the 2nd number system end in 0 - even, on 1 - odd;
• respectively, numbers that are divisible by 4 , will end in 00 , etc.; thus we derive general rule: numbers that are divisible by 2 k, ending in k zeros
• if number N is in the interval 2 k-1 ≤ N , in its binary notation there will be exactly k numbers, for example 126 :

2 6 = 64 ≤ 126

if the number looks like 2 k, then it is written in binary as unit and k zeros, for example:

32 = 2 5 = 100000 2

if the number looks like 2k-1, then it is written in the binary system k units, for example:

31 = 2 5 -1 = 11111 2

if the binary notation is known N, then the binary notation of the number 2 N can be easily obtained by appending zero to the end, for example:

15 = 1111 2 , 30 = 11110 2 , 60 = 111100 2 , 120 = 1111000 2

You also need to learn powers of two, increasing the degree from right to left:

1024 512 256 128 64 32 16 8 4 2 1 2 10 2 9 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0

it is desirable to learn a table of binary representation of numbers from 0 to 7 in the form of triads (groups of 3 bits):

X 10 ,X 8 X 2 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111

it is desirable to know the table of binary representation of numbers from 0 to 15 (in hexadecimal s-me - 0-F 16) in the form of tetrads (groups of 4 bits):

X 10 X 16 X 2 0 0 0000 1 1 0001 2 2 0010 3 3 0011 4 4 0100 5 5 0101 6 6 0110 7 7 0111 8 8 1000 9 9 1001 10 A 1010 11 B 1011 10 4 C 1 10 1110 15 F 1111

Translation of negative ( -a) to two's complement is done as follows:

• need to be translated a-1 to the binary number system;
• do a bit inversion: change all zeros to ones and ones to zeros within the bit grid

Solution 1 task of the exam

1_1: Parsing 1 USE assignments in Informatics 2017 FIPI option 1 (Krylov S.S., Churkina T.E.):

How units in binary hexadecimal notation 2AC1 16?

✍ Solution:

• In hexadecimal notation, numbers from 10 before 15 represented by letters of the Latin alphabet: A-10, B-11, C-12, D-13, E-14, F-15.
• You need to remember the binary codes of numbers from 1 before 15 (see the theory above on the page), since to convert hexadecimal to binary s-mu, it is enough to write each digit separately as a four of binary digits (tetrads):

2 A C 1 0010 1010 1100 0001

there are 6 units in this post

Result: 6

Detailed analysis of 1 task with an explanation, see the video:

1_2: 1 task. Demo version of the exam 2018 informatics (FIPI):

How many integers are there x, for which the inequality 2A 16<x<61 8 ?
Answer only the number of numbers.

✍ Solution:

• Let's translate 2A 16 into the decimal number system:

2A 16 = 2*16 1 +10*16 0 = 32 + 10 = 42

Let's translate 61 8 into decimal sth notation:

61 8 = 6*8 1 +1*8 0 = 48 + 1 = 49

Let's get a comparison:

42

Since the task has a double strict comparison (<), то количество целых, удовлетворяющих условию:

49 - 42 - 1 = 6

Let's check: 43, 44, 45, 46, 47, 48

Result: 6

For a detailed solution of this 1 task from the USE demo version of 2018, see the video:

1_3: 1 task. GVE Grade 11 in Informatics 2018 (FIPI):

How many significant digits are in the binary notation of a decimal number 129 ?
1) 6
2) 2
3) 7
4) 8

✍ Solution:

• Let's translate from decimal s-we reckoning to binary by dividing by 2 , on the right we will write the remainders:

129 / 1 64 / 0 32 / 0 16 / 0 8 / 0 4 / 0 2 / 0 1

Let's rewrite the remainders from bottom to top, starting with the last unit, which is no longer divisible by two:

10000001

Let's count the number of digits in the resulting binary number. There are 8 of them, and they are all significant (only zeros on the left can be insignificant, for example, 010 is the same as 10 ). The correct answer is number 4

Result: 4

1_4: Solution 1 task of the Unified State Examination in Informatics (control version of the examination paper in 2018, S.S. Krylov, D.M. Ushakov):

How many natural numbers x are there for which the inequality

101011 2 ?

Answer only the number of numbers.

✍ Solution:

Result: 17

We offer a detailed analysis of the solution of the training option in the video:

1_5: Analysis of 1 task of the Unified State Examination option No. 1, 2019 Informatics and ICT Typical exam options (10 options), S.S. Krylov, T.E. Churkina::

Compute expression value AE 16 – 19 16.
In your answer, write down the calculated value in decimal notation.

✍ Solution:

• Let's translate the minuend and the subtrahend into the decimal number system:

1 0 A E = 10*16 1 + 14*16 0 = 160 + 14 = 174

* A 16 corresponds to the number 10 in decimal notation

* E 16 corresponds to the number 14 in decimal notation

1 0 19 = 1*16 1 + 9*16 0 = 16 + 9 = 25

Let's find the difference:

174 - 25 = 149

Result: 149

1_6: Analysis of task 1 of the exam (from the site of K. Polyakov, option 104 with reference to Noskin A.N.):

Petya and Kolya think of natural numbers. Peter guessed the number X, and Kohl's number At. After Petya added to Kolya's number 9 , and Kolya to Petya's number 20 , the sum of the obtained numbers when written in the binary system is five units.

What is equal to initial amount numbers guessed by the boys? Write your answer in binary number system. The reason does not need to be stated.

✍ Solution:

• Let's rewrite the condition of the problem in a more understandable form:

(x + 9) + (y + 20) = 11111 2 (x + y) 2 = ?

Let's translate 11111 2 into the decimal number system and subtract from the result the numbers of Kolya and Petya to get just the sum (x+y):

11111 2 = 31 10 31 - 20 - 9 = 2

Let's translate the result into the binary number system:

2 10 = 10 2

Result: 10

1_7: Analysis of task 1 of the exam (from the site of K. Polyakov, option 105 with reference to Kutsyr E.V.):

Specify largest four-digit octal number, quaternary whose entry contains exactly 2 triplets, not standing next to. In your answer, write down only the octal number itself, you do not need to indicate the base of the number system.

✍ Solution:

• Recall that in the octal system the maximum number is 7, and in the quaternary system - 3. Let's try to convert the largest octal number to the quaternary system, not taking into account the condition with non-consecutive triples. Let's translate through the binary number system:

7777 8 - maximum four-digit octal number Convert to binary s.s.: 7 7 7 7 111 111 111 111 Convert from binary s.s. in the quaternary is carried out by dividing into groups of two digits: 11 11 11 11 11 11 3 3 3 3 3 3

Thus, in order to get the largest quaternary number containing two triples that are not in a row, you need to remove one unit from all groups in its binary notation, except for two that belong to the highest digits and are not in a row:

11 10 11 10 10 10 3 2 3 2 2 2 4

Let's translate the result into the 8th number system:

111 011 101 010 7 3 5 2

Result: 7352

1_8: Analysis of task 1 of the exam (from the site of K. Polyakov, option 109 with reference to Noskin A.N.):

Segment is set . Number a- the smallest number octal whose entry contains exactly 3 characters, one of which is 3 . Number b – least number, hexadecimal whose record contains exactly 3 characters, one of which is F.

Determine number of natural numbers on this segment ( including its ends).

✍ Solution:

• Let's rewrite the condition of the problem in a more understandable form, substituting the values ​​for the numbers a and b:

a: 103 8 - the smallest three-digit octal number, one of the digits of which is - 3 b: 10F 16 - the smallest three-digit 16th number, one of the digits of which is - F

Let's translate the numbers into the decimal number system and find the length of the segment by performing the difference of these numbers:

103 8 \u003d 67 10 10F 16 \u003d 271 10 \u003d length of the segment \u003d 271 - 67 + 1 (including its ends) \u003d 205

Result: 205

1_9: Solution 1 task of the USE 2020 (Thematic training tasks, 2020, Samylkina N.N., Sinitskaya I.V., Soboleva V.V.):

One byte is used to store a signed integer.

How many units contains the internal representation of a number ( -116 )?

✍ Solution:

To convert a negative number to a binary number system, we use the following algorithm:
• Subtract one from the modulus of the original number:

|-116| - 1 = 115

Let's translate the result into a binary number system:

115 10 = 1110011 2

Since one byte is used for storage, it is necessary to supplement the resulting number with insignificant zeros on the left to 8 digits:

0 1110011

Invert the result (replace ones with zeros, and zeros with ones):

10001100

Result: 10001100

With the modern world of technology and the realities of programming, development USE in Informatics has little in common. There are some basic points, but even if you understand tasks a little, this does not mean that you will eventually become a good developer. But there are a lot of areas where IT specialists are needed. You will not lose at all if you want to have a stable income above the average. In IT, you get it. Provided, of course, that you have the appropriate skills. And you can develop and grow here as much as you like, because the market is so huge that you can’t even imagine! And it is not limited only to our state. Work for any company from anywhere in the world! This is all very inspiring, so let the preparation for the exam in computer science be the first small step, after which years of self-development and improvement in this area will follow.

Structure

Part 1 contains 23 short answer tasks. This part contains tasks with a short answer, implying an independent formulation of a sequence of characters. Tasks check the material of all thematic blocks. 12 tasks belong to the basic level, 10 tasks to an increased level of complexity, 1 task to a high level of complexity.

Part 2 contains 4 tasks, the first of which is of an increased level of complexity, the remaining 3 tasks are of a high level of complexity. The tasks of this part involve writing a detailed answer in an arbitrary form.

3 hours 55 minutes (235 minutes) are allotted to complete the examination paper. It is recommended to take 1.5 hours (90 minutes) to complete the tasks of part 1. The rest of the time is recommended to be devoted to the tasks of part 2.

Explanations for grading assignments

The performance of each task of part 1 is estimated at 1 point. The task of part 1 is considered completed if the examiner gave the answer corresponding to the code of the correct answer. Completion of the tasks of part 2 is estimated from 0 to 4 points. The answers to the tasks of part 2 are checked and evaluated by experts. The maximum number of points that can be obtained for completing the tasks of part 2 is 12.