Schoolchildren from the Khabarovsk Territory opened the Unified State Exam season by passing computer science and geography (photo; video). Collection of ideal social studies essays

The busy season has begun for schoolchildren in the Khabarovsk Territory - they have begun taking the Unified State Exam. The first were computer science and geography. In the Khabarovsk Territory, more than 700 people took these items. Before the start of the test, the guys went through metal detectors and passport control.

Last week the last bells rang for school graduates, and today it’s time for exams. Computer science and geography are optional subjects, but hundreds of people volunteered to take them.

I was always interested in computers and systems, I even wanted to become a programmer, so I decided to take this subject. I solved many variants of the Unified State Exam and standard tasks. There is some excitement, but I have already come to terms with it. The most difficult part for me is the last part. There are several types of programs that need to be written, and this has to be done just on a piece of paper, said Daniil Kharkov, a graduate of the Vector school.

Among those gathered to take computer science were great amount girls. The graduates prepared for the exam not only intellectually, but also with the help of signs. So, one girl reported about pens for each subject of the Unified State Examination, another about a heel under the heel. There were also those who believed more in themselves than in any rituals.

I was very lucky at school there was a very good and demanding teacher, thanks to which there was no need to turn to tutors. I started preparing for the Unified State Exam three months ago. I don’t really believe in omens, but I took the advice. On " last call“We were advised not to wash our hair before the exam, so as not to wash the knowledge out of our heads. I chose this subject because it is easy for me. I take it more to be on the safe side, since the second chosen discipline is physics,” said Ekaterina Maksimova, a school graduate.

The exam began at 10 a.m., but before that, students were checked through a metal detector, subjected to body searches, and escorted to their classrooms. Everything was done slowly, but carefully, so that no incidents would arise later.

Computer science is given 3 hours 55 minutes, geography exactly three hours. We have students with disabilities, they have the right to extend the exam by one and a half hours. We have been observing for several years that the proportion of schoolchildren who chose these subjects has not changed. These disciplines are not as in demand by universities as social science or physics. And computer science and geography are a test for rare specialties, said Ekaterina Koshelnikova, head of the department of final certification and assessment of the quality of education of the department of the Ministry of Education and Science of the region.

It should be noted that 575 people in the region write computer science, 149 people write geography. You cannot use any additional materials or tools in computer science. In geography, everything is a little simpler - you are allowed to bring a ruler, protractor, and calculator. The exam results will be known on June 13.

Let us remind you that ninth grade schoolchildren and their parents will be able to find out the results of the compulsory state exam (OGE) online. New opportunity provided on the regional government services portal. To receive it, the student just needs to go to electronic service and enter your last name, first name, patronymic and passport number.

WITH modern world technologies and realities of programming, development Unified State Exam in Computer Science has little in common. There are some basic points, but even if you understand a little about the tasks, this does not mean that you will ultimately become a good developer. But there are a great many areas where IT specialists are needed. You can't go wrong if you want to have a stable income above average. In IT you will get it. Provided, of course, that you have the appropriate abilities. And you can develop and grow here as much as you want, because the market is so huge that you can’t even imagine! Moreover, it is not limited only to our state. Work for any company from anywhere in the world! This is all very inspiring, so let preparation for the Unified State Exam in computer science be the first minor step, followed by years of self-development and improvement in this area.

Structure

Part 1 contains 23 short answer questions. This part contains short-answer tasks that require you to independently formulate a sequence of symbols. The assignments test 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 difficulty, the remaining 3 tasks high level difficulties. The tasks in this part involve writing a detailed answer in free form.

3 hours 55 minutes (235 minutes) are allotted to complete the examination work. It is recommended to spend 1.5 hours (90 minutes) to complete the tasks of Part 1. It is recommended to devote the rest of the time to completing the tasks of part 2.

Explanations for grading assignments

Completion of each task in Part 1 is worth 1 point. Part 1 task is considered completed if the examinee gives an answer that corresponds to the correct answer code. Completion of tasks in part 2 is graded from 0 to 4 points. Answers to tasks in Part 2 are checked and assessed by experts. Maximum amount The points you can get for completing tasks in part 2 are 12.

The lesson discusses solution 12 Unified State Exam assignments in computer science, including assignments for 2017

Topic 12 - “ Network addresses» - characterized as tasks basic level complexity, execution time - approximately 2 minutes, maximum score - 1

Internet addressing

The address of a document on the Internet (from English - URL - Uniform Resource Locator) consists of the following parts:

• data transfer protocol; May be:
• http(for Web pages) or
• ftp(for file transfer)
• there is also a secure protocol https;
• delimiter characters :// , separating the protocol name from the rest of the address;
• Domain name website (or IP address);
• may also be present: the directory on the server where the file is located;
• file name.

Directories on the server are separated by forward slash " / »

1. network service protocol name – defines the server type HTTP(Hypertext Transfer Protocol);
2. delimiter in the form of a colon and two characters Slash;
3. fully qualified domain name of the server;
4. search path for a web document on a computer;
5. web server name;
6. top level domain "org";
7. Name national domain "ru";
8. catalog main on the computer;
9. catalog news in the catalog main;
10. the final goal of the search is a file main_news.html.

Network addresses

Physical adress or MAC address– unique address, “hardwired” in production – 48-bit code network card(in hexadecimal):

00-17-E1-41-AD-73

IP address– computer address (32-bit number), consisting of: network number + computer number in the network (node ​​address):

15.30.47.48

Subnet mask:

• necessary to determine which computers are on the same subnet;

in the 10th performance in the 16th performance

255.255.255.0 -> FF.FF.FF.0

mask in binary code always has the structure: first all ones, then all zeros:

1…10…0

when superimposed on an IP address (logical conjunction AND) gives the network number:

The part of the IP address that corresponds to the mask bits equal to one refers to the network address, and the part corresponding to the mask bits equal to zero is the numerical address of the computer

thus, it is possible to determine what it may be last number of mask:

if two nodes belong to the same network, then their network address is the same.

Calculation of network number by IP address and network mask

In the subnet mask most significant bits, allocated in the computer's IP address for network number, have a value of 1 (255); least significant bits, allocated in the computer's IP address for computer addresses in the subnet, matter 0 .

* Image taken from the presentation by K. Polyakov

Number of computers on the network

The number of computers on the network is determined by the mask: the low-order bits of the mask - zeros - are reserved in the computer's IP address for the address of the computer in the subnet.

If the mask:

The number of computers on the network:

2 7 = 128 addresses

Of these, 2 are special: network address and broadcast address

128 - 2 = 126 addresses

Solving tasks 12 Unified State Exam in computer science

Unified State Examination in Informatics 2017 task 12 FIPI option 1 (Krylov S.S., Churkina T.E.):

In TCP/IP network terminology, a network mask is called binary number, which determines which part of the IP address of a network host refers to the network address, and which part refers to the address of the host itself on this network. Typically, the mask is written according to the same rules as the IP address - as four bytes, with each byte written as a decimal number. In this case, the mask first contains ones (in the highest digits), and then from a certain digit there are zeros. The network address is obtained by applying a bitwise conjunction to the given host IP address and mask.

For example, if the host IP address is 211.132.255.41 and the mask is 255.255.201.0, then the network address is 211.132.201.0

For a node with an IP address 200.15.70.23 network address is 200.15.64.0 . What is equal to least possible value of the third byte from the left of the mask? Write your answer as a decimal number.

✍ Solution:

• The third byte from the left corresponds to the number 70 in the IP address and 64 - in the network address.
• The network address is the result of the bitwise conjunction of the mask and the IP address in binary:

? ? ? ? ? ? ? ? -> third byte of the mask AND (&) 0 1 0 0 0 1 1 0 2 -> 70 10 = 0 1 0 0 0 0 0 0 2 -> 64 10

The smallest possible result of the mask could be:

1 1 0 0 0 0 0 0 - third byte of the mask AND (&) 0 1 0 0 0 1 1 0 2 -> 70 10 = 0 1 0 0 0 0 0 0 2 -> 64 10

Here the most significant bit is taken as one, although the result of the conjunction could have been taken as zero (0 & 0 = 0). However, since there is a guaranteed one next to it, it means that we also put it in the most significant bit 1 . As you know, the mask contains ones first, and then zeros (this cannot happen: 0100… , but it can only be like this: 1100… ).

Let's translate 11000000 2 into the 10th number system and we get 192 .

Result: 192

A step-by-step solution to this 12th task of the Unified State Exam in computer science is available in the video tutorial:

Task 12. Demo version of the Unified State Exam 2018 computer science:

In the terminology of TCP/IP networks, a network mask is a binary number that determines which part of the IP address of a network host refers to the network address, and which part refers to the address of the host itself on this network. Usually the mask is written according to the same rules as the IP address - in as four bytes, with each byte written as a decimal number. In this case, the mask first contains ones (in the highest digits), and then from a certain digit there are zeros.
The network address is obtained by applying a bitwise conjunction to the given 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 node with an IP address 57.179.208.27 network address is 57.179.192.0 . What's it like greatest possible quantity units in the ranks of the mask?

✍ Solution:

• Since the network address is obtained as a result of applying a bitwise conjunction to a given host IP address and mask, we get:

255.255.?.? -> mask & 57.179.208.27 -> IP address = 57.179.192.0 -> network address

Since the first two bytes on the left in the host IP address and the network address are the same, it means that in order to obtain such a result in a bitwise conjunction in the binary system, the mask must contain all ones. Those.:

11111111 2 = 255 10

In order to find the remaining two bytes of the mask, it is necessary to convert the corresponding bytes in the IP address and network address to the 2nd number system. Let's do it:

208 10 = 11010000 2 192 10 = 11000000 2

Now let's see what the mask for this byte can be. Let's number the bits of the mask from right to left:

7 6 5 4 3 2 1 0 1 1 1 0 0 0 0 0 -> mask & 1 1 0 1 0 0 0 0 = 1 1 0 0 0 0 0 0

For the 5th bit we get: ? & 0 = 0 -> the mask can contain both a unit and 0 . But since the assignment asks us greatest possible number of units, which means it is necessary to say that the mask is given bit equals 1 .

For the 4th bit we get: ? & 1 = 0 -> the mask can only be worn by 0 .

Since the mask contains first ones and then all zeros, then after this zero in the 4th bit all the rest will be zeros. And the 4th byte from the left of the mask will be equal to 0 10 .

Let's get the mask: 11111111.11111111.11100000.00000000 .

Let's count the number of units in the mask:

8 + 8 + 3 = 19

Result: 19

For a detailed solution to task 12 of the demo version of the Unified State Exam 2018, watch the video:

Solution to task 12 (Polyakov K., option 25):

In TCP/IP network terminology, a network mask is a binary number that indicates which part of the IP address of a network host relates to the network address, and which part to the host address on this network. The network address is obtained by applying a bitwise conjunction to a given node address and its mask.

Based on the specified host IP address and mask determine the network address:

IP address: 145.92.137.88 Mask: 255.255.240.0

When recording the response, select the four elements of the IP address from the numbers given in the table and write in in the right order their corresponding letters without dots.

A	B	C	D	E	F	G	H
0	145	255	137	128	240	88	92

✍ Solution:

• To solve the problem, you need to remember that the IP address of the network, as well as the network mask, are stored in 4 bytes written with a dot. That is, each of the individual IP address and netmask numbers are stored in 8-bit binary form. To obtain the network address, it is necessary to perform a bitwise conjunction of these numbers.
• Since the number 255 in binary representation this is 8 units, then with a bitwise conjunction with any number, the result will be the same number. Thus, there is no need to take into account those bytes of the IP address that correspond to the number 255 in the network mask. Therefore, the first two numbers of the IP address will remain the same ( 145.92 ).
• It remains to consider the numbers 137 And 88 IP addresses and 240 masks. Number 0 in the mask matches eight zeros in binary representation, that is, a bitwise conjunction with any number will turn this number into 0 .
• Let's convert both numbers of the IP address and network mask to binary system and write the IP address and mask under each other to implement the bitwise conjunction:

137: 10001001 88: 1011000 - IP address 240: 11110000 0: 00000000 - network mask 10000000 00000000 - the result of the bitwise conjunction

Let's translate the result:

10000000 2 = 128 10

In total, for the network address we get the bytes:

145.92.128.0

We match the letters in the table and get BHEA.

Result: BHEA

We invite you to watch a detailed video analysis:

Solution to task 12 (Polyakov K., option 33):

If the subnet mask 255.255.255.128 and IP address of the computer on the network 122.191.12.189 , then the computer number on the network is _____.

✍ Solution:

• Unit mask bits ( equal to one) determine the subnet address, because The subnet address is the result of the bitwise conjunction (logical multiplication) of the mask bits with the IP address.
• The rest of the mask (starting with the first zero) specifies the computer number.
• Since in binary representation the number 255 - this is eight units ( 11111111 ), then with a bitwise conjunction with any number, the same number is returned (1 ∧ 0 = 0; 1 ∧ 1 = 1). Thus, those bytes in the mask that are equal to numbers 255 , we will not consider, because they define the subnet address.
• Let's start with a byte equal to 128 . It corresponds to a byte 189 IP addresses. Let's convert these numbers to the binary number system:

128 = 10000000 2 189 = 10111101 2

Those bits of the IP address that correspond to the zero bits of the mask are used to determine the computer number. Let's convert the resulting binary number to decimal system notation:

0111101 2 = 61 10

Result: 61

Detailed solution of this assignment look at the video:

Solution to task 12 (Polyakov K., option 41):

In the terminology of TCP/IP networks, a subnet mask is a 32-bit binary number that determines which bits of the computer's IP address are common to the entire subnet - these bits of the mask contain 1. Masks are usually written in the form of a quadruple decimal numbers- according to the same rules as IP addresses.

A mask is used for some subnet 255.255.255.192 . How many different computer addresses theoretically allows this mask if two addresses (network and broadcast address) are not used?

✍ Solution:

• The single bits of the mask (equal to one) determine the subnet address, the rest of the mask (starting with the first zero) determines the computer number. That is, there are as many options for the computer address as can be obtained from the zero bits in the mask.
• In our case, we will not consider the first three bytes of the mask on the left, because number 255 in binary representation it is eight units ( 11111111 ).
• Consider the last byte of the mask, equal to 192 . Let's convert the number to the binary number system:

192 10 = 11000000 2

Total received 6 zeros in the network mask. This means that 6 bits are allocated for addressing computers or, in other words, 2 6 computer addresses. But since two addresses are already reserved (by condition), we get:

2 6 - 2 = 64 - 2 = 62

Result: 62

Watch the video description of the task below:

Solution to task 12 (Edge work, Far East, 2018):

For a node with an IP address 93.138.161.94 network address is 93.138.160.0 .For how many different meanings masks is this possible?

✍ Solution:

Result: 5

Video analysis of the task: