Вспомни общие закономерности набора символов. нажатие кнопки английская раскладка нажатие кнопки + Shift русская раскладка 1) В каком ряду кнопки нужный символ? Верхний. Нижний 2) В каком столбце кнопки нужный символ" Левый Правый

Var
A,B0,B1,B2,B3,B4,B5,B6,B7,B8,B9:integer;
Begin
WriteLn('Введите число');
ReadLn(A);
While A>0 do
   Begin
   Case A mod 10 of
   0:B0:=B0+1;
   1:B1:=B1+1;
   2:B2:=B2+1;
   3:B3:=B3+1;
   4:B4:=B4+1;
   5:B5:=B5+1;
   6:B6:=B6+1;
   7:B7:=B7+1;
   8:B8:=B8+1;
   9:B9:=B9+1;
   End;
   A:= A div 10;
   End;
if (B0 = 3)and(B1<2)and(B2<2)and(B3<2)and(B4<2)and(B5<2)and(B6<2)and(B7<2)and(B8<2)and(B9<2)then WriteLn('Верно')
else if (B1 = 3)and(B2<2)and(B3<2)and(B4<2)and(B5<2)and(B6<2)and(B7<2)and(B8<2)and(B9<2)and(B0<2)then WriteLn('Верно')
else if (B2 = 3)and(B3<2)and(B4<2)and(B5<2)and(B6<2)and(B7<2)and(B8<2)and(B9<2)and(B0<2)and(B1<2)then WriteLn('Верно')
else if (B3 = 3)and(B4<2)and(B5<2)and(B6<2)and(B7<2)and(B8<2)and(B9<2)and(B0<2)and(B1<2)and(B2<2)then WriteLn('Верно')
else if (B4 = 3)and(B5<2)and(B6<2)and(B7<2)and(B8<2)and(B9<2)and(B0<2)and(B1<2)and(B2<2)and(B3<2)then WriteLn('Верно')
else if (B5 = 3)and(B6<2)and(B7<2)and(B8<2)and(B9<2)and(B0<2)and(B1<2)and(B2<2)and(B3<2)and(B4<2)then WriteLn('Верно')
else if (B6 = 3)and(B7<2)and(B8<2)and(B9<2)and(B0<2)and(B1<2)and(B2<2)and(B3<2)and(B4<2)and(B5<2)then WriteLn('Верно')
else if (B7 = 3)and(B8<2)and(B9<2)and(B0<2)and(B1<2)and(B2<2)and(B3<2)and(B4<2)and(B5<2)and(B6<2)then WriteLn('Верно')
else if (B8 = 3)and(B9<2)and(B0<2)and(B1<2)and(B2<2)and(B3<2)and(B4<2)and(B5<2)and(B6<2)and(B7<2)then WriteLn('Верно')
else if (B9 = 3)and(B0<2)and(B1<2)and(B2<2)and(B3<2)and(B4<2)and(B5<2)and(B6<2)and(B7<2)and(B8<2)then WriteLn('Верно')
else WriteLn('Не верно')
End.

#include <iostream>

int main() {

  int arr[100], n, first = 0, second = 0;

  std::cin >> n;

  for (int i = 0; i < n; ++i)

      std::cin >> arr[i];

  for (int i = 1; i < n; ++i) {

      const int diff = arr[i] - arr[i - 1];

      if (diff > 0)

          if (arr[i] <= 2000)

              first += diff;

          else if (arr[i - 1] >= 2000)

              second += diff;

          else {

              first += 2000 - arr[i - 1];

              second += arr[i] - 2000;

           }

   }

  std::cout << first << ' ' << second;

  return 0;

Объяснение:

язык python