题解 | #压缩二维码#（Python3）

import sys

def func(n, arr):
    # 矩阵转字符串
    s = ''.join(arr)
    ans = []
    # 4个比特位算一个整数
    for i in range(0, n*n, 4):
        w = 0
        # 计算整数值
        for k in range(0, 4):
            w = w * 2 + (1 if s[i+k]== '#' else 0)
        ans.append(w)
    return ans

# 输入读取
while True:
    try:
        n = int(input())
        n = 1<<n
        arr = [input() for i in range(0, n)]
        result = func(n, arr)
        for number in result:
            print(number, end=' ')
        print()
    except EOFError:
        break