1416- Find the Lost Sock(异或运算) ZCMU

Description

Alice bought a lot of pairs of socks yesterday. But when she went home, she found that she has lost one of them. Each sock has a name which contains exactly 7 charaters.

Alice wants to know which sock she has lost. Maybe you can help her.

Input

There are multiple cases. The first line containing an integer n (1 <= n <= 1000000) indicates that Alice bought n pairs of socks. For the following 2*n-1 lines, each line  is a string with 7 charaters indicating the name of the socks that Alice took back.

Output

The name of the lost sock.

Sample Input

2

aabcdef

bzyxwvu

bzyxwvu

4

aqwerty

easafgh

aqwerty

easdfgh

easdfgh

aqwerty

aqwerty

2

0x0abcd

0ABCDEF

0x0abcd

Sample Output

aabcdef

easafgh

0ABCDEF

解析

看到这道题的第一反应是用map做,但发现超时。看了别人的代码,发现还有更简单的方法-异或运算

代码

#include<stdio.h>
int main()
{
    int n,i,j;
    char s[10],sock[10];
    while(~scanf("%d",&n))
    {
        scanf("%s",s);
        for(i=2;i<=2*n-1;i++)
        {
            scanf("%s",sock);
            for(j=0;j<7;j++)
            {
                s[j]^=sock[j];
            }
        }
        printf("%s\n",s);
    }
    return 0;
}
// A^B^A=A^(B^A)=A^A^B=B

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值