nq1s788

перебор всех квадратов 2 на 2

Nov 2nd, 2025
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Python 0.40 KB | None | 0 0
  1. n = 4
  2. a = [[1, 0, 0, 0],
  3.      [1, 1, 0, 1],
  4.      [1, 1, 1, 1],
  5.      [1, 0, 1, 1]]
  6. for i in range(1, n):
  7.      for j in range(1, n):
  8.           if a[i - 1][j - 1] == a[i - 1][j] == a[i][j - 1] == a[i][j] == 0:
  9.                print('квадрат с нулями')
  10.           if a[i - 1][j - 1] == a[i - 1][j] == a[i][j - 1] == a[i][j] == 1:
  11.                print('квадрат с единицами')
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