
这题其实没什么难度,主要就是要读取文件,整理成一个矩阵,然后遍历整个矩阵,每个是2的位置判断三个方向是否构成2020就可以了。
Code Pythonif __name__ == '__main__':
matrix = []
with open("./2020.txt", 'r') as fp:
for line in fp.readlines():
line = line.strip()
matrix.append(list(line))
ans = 0
for i in range(len(matrix)):
for j in range(len(matrix[i])):
if matrix[i][j] == '2':
if j + 3 < len(matrix[i])
and matrix[i][j + 1] == '0'
and matrix[i][j + 2] == '2'
and matrix[i][j + 3] == '0':
ans += 1
if i + 3 < len(matrix)
and matrix[i + 1][j] == '0'
and matrix[i + 2][j] == '2'
and matrix[i + 3][j] == '0':
ans += 1
if i + 3 < len(matrix)
and j + 3 < len(matrix[i])
and matrix[i + 1][j + 1] == '0'
and matrix[i + 2][j + 2] == '2'
and matrix[i + 3][j + 3] == '0':
ans += 1
print(ans)
Answer:16520欢迎分享,转载请注明来源:内存溢出
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