Question

Var a, b, с : integer; begin a := 0; b := 0; с := 0; while 2 * a < 200 do begin b := b + 3; c := c - 1; a := a + b + c end; write(a - 10) end.ЧТО ПОЛУЧИТСЯ В ОТВЕТЕ??????СРОЧНО,ПОЖАЛУЙСТА,ПОМОГИТЕ!!!!

Answer 1

Ответ, полученный в Pascal ABC: 100.

Answer 2

Code Explanation:

The given code is a Pascal program that initializes three variables `a`, `b`, and `c` as integers and then enters a while loop. The loop continues as long as `2 * a` is less than 200. Inside the loop, the values of `b` and `c` are updated, and `a` is incremented by the sum of `b` and `c`. Finally, the program outputs the value of `a - 10`.

Let's go through the code step by step to understand what the final output will be.

1. Initialize variables: - `a` is set to 0. - `b` is set to 0. - `c` is set to 0.

2. Enter the while loop: - The loop condition is `2 * a < 200`, which means the loop will continue as long as `2 * a` is less than 200.

3. Inside the loop: - `b` is incremented by 3 (`b := b + 3`). - `c` is decremented by 1 (`c := c - 1`). - `a` is incremented by the sum of `b` and `c` (`a := a + b + c`).

4. Repeat steps 2 and 3 until the loop condition is false.

5. After the loop: - The program outputs the value of `a - 10`.

Now, let's execute the code and determine the final output.

Code Execution:

1. Initialize variables: - `a = 0` - `b = 0` - `c = 0`

2. Enter the while loop: - Since `2 * a = 2 * 0 = 0`, which is less than 200, the loop condition is true.

3. Inside the loop: - `b` is incremented by 3: `b = 0 + 3 = 3` - `c` is decremented by 1: `c = 0 - 1 = -1` - `a` is incremented by the sum of `b` and `c`: `a = 0 + 3 + (-1) = 2`

4. Repeat steps 2 and 3: - `2 * a = 2 * 2 = 4`, which is still less than 200, so the loop continues.

- `b` is incremented by 3: `b = 3 + 3 = 6` - `c` is decremented by 1: `c = -1 - 1 = -2` - `a` is incremented by the sum of `b` and `c`: `a = 2 + 6 + (-2) = 6`

- `2 * a = 2 * 6 = 12`, which is still less than 200, so the loop continues.

- `b` is incremented by 3: `b = 6 + 3 = 9` - `c` is decremented by 1: `c = -2 - 1 = -3` - `a` is incremented by the sum of `b` and `c`: `a = 6 + 9 + (-3) = 12`

- This process continues until `2 * a` is no longer less than 200.

5. After the loop: - The final value of `a` is 12. - The program outputs `a - 10`: `12 - 10 = 2`.

Answer:

The final output of the given code will be 2.