Question

S:=0; for i:=1 to 3 do begin s:=s+1; end; write (s); 1 3 2 4​

Answer 1

Ответ:

Program zikl;

Var s, i:Integer;

Begin

s:=0;

for i:=1 to 3 do

begin

s:=s+1;

Writeln(s)

end;

End.

Выводит 1, 2, 3.

Если нужно до 4, то замени строку for i:=1 to 3 do, строкой: for i:=1 to 4 do

Answer 2

The code snippet you provided is a simple loop that iterates from 1 to 3. In each iteration, the variable "s" is incremented by 1. Finally, the value of "s" is displayed using the "write" command.

Here's how the code executes:

1. Initialize the variable "s" to 0: S := 0.
2. Start the loop: for i := 1 to 3 do begin.
3. In the first iteration, i is 1. Increment "s" by 1: S := S + 1. The value of "s" is now 1.
4. In the second iteration, i is 2. Increment "s" by 1 again: S := S + 1. The value of "s" is now 2.
5. In the third iteration, i is 3. Increment "s" by 1: S := S + 1. The value of "s" is now 3.
6. The loop ends because the condition (i <= 3) is no longer true.
7. Display the final value of "s" using the "write" command: write(s). The output will be 3.

Therefore, the correct answer is 3.