Question

Массив задается, а затем меняется: for i := 1 to 10 do a[i] := i + 1; for i := 1 to 9 do begin tmp := a[i + 1]; a[i + 1] := a[i]; a[i] := tmp; end; Как выглядит массив после выполнения программы? 1) 3, 2, 5, 4, 7, 6, 9, 8, 11, 10 2) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 3) 11, 10, 9, 8, 7, 6, 5, 4, 3, 2 4) 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 5) 3, 4, 5, 6, 7, 8, 9, 10, 11, 2

Answer 1

Правильный вариант ответа: 5.

Answer 2

To determine the array values after executing the given code, let's simulate the operations step by step:

1. Initialization of the array: a = [2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

2. Swap operation loop (for i := 1 to 9 do ...): i = 1: Swap a[2] and a[1]: a = [3, 2, 4, 5, 6, 7, 8, 9, 10, 11] i = 2: Swap a[3] and a[2]: a = [3, 4, 2, 5, 6, 7, 8, 9, 10, 11] i = 3: Swap a[4] and a[3]: a = [3, 4, 5, 2, 6, 7, 8, 9, 10, 11] i = 4: Swap a[5] and a[4]: a = [3, 4, 5, 6, 2, 7, 8, 9, 10, 11] i = 5: Swap a[6] and a[5]: a = [3, 4, 5, 6, 7, 2, 8, 9, 10, 11] i = 6: Swap a[7] and a[6]: a = [3, 4, 5, 6, 7, 8, 2, 9, 10, 11] i = 7: Swap a[8] and a[7]: a = [3, 4, 5, 6, 7, 8, 9, 2, 10, 11] i = 8: Swap a[9] and a[8]: a = [3, 4, 5, 6, 7, 8, 9, 10, 2, 11]

After executing the code, the array 'a' will be: 3, 4, 5, 6, 7, 8, 9, 10, 2, 11

So, the correct option is: 5) 3, 4, 5, 6, 7, 8, 9, 10, 2, 11