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
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