借助掩码将face映射成一个二进制数,如果字符的二进制数相同,则字符相同 n,m=map(int,input().split()) t=0 g=[] for i in range(n): g.append(list(input())) for x in 'face': t|=1<<(ord(x)-97) r=0 for i in range(n-1): for j in range(m-1): cur=0 for x in [g[i][j],g[i+1][j],g[i][j+1],g[i+1][j+1]]: cur |= 1<< ord(x)-97 if cur==t:...