E题题解 | #井字棋#
井字棋
https://ac.nowcoder.com/acm/contest/98530/E
E题的妙妙解法
题目确保了棋局已经结束。则若先手赢，先手下的棋子比后手多。则若后手赢，先手下的棋子和后手一样多。
代码如下
maze = [list(input()) for i in range(3)]
cnt = [0, 0]
for i in range(3):
for j in range(3):
cnt[0] += (maze[i][j] == 'X')
cnt[1] += (maze[i][j] == 'O')
if cnt[0] > cnt[1]:
print('Qfish win!')
else:
print("lumo win!")
