描述

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.

輸入

There are multiple cases. The first line containing an integer n

輸出

The name of the lost sock.

樣例輸入

2
aabcdef
bzyxwvu
bzyxwvu
4
aqwerty
eas fgh
aqwerty
easdfgh
easdfgh
aqwerty

樣例輸出

aabcdef
eas fgh
0ABCDEF

提示

Because of HUGE input, scanf is recommended.

解題思路：

數據量很大，用map存會超時。由於一位學長的指點，了解到了按位異或，當一個數出現了兩次的時候就會抵消，所以我們只要不斷按位異或，最後剩下的就是答案。

比如 1 1 3 2 2

我們只要按位異或 1->0->3->1->3

最後的3便是所求的答案。

#include <bits/stdc++.h>
using namespace
 
 std;
int main()
{
    int n,m,i,j,k;
    while(cin>>n)
    {
        getchar();
        int s[8]={0};char a[10];
        for(i=1;i<=2*n-1;i++)
        {
            gets(a);
            for(j=0;j<7;j++) s[j]=s[j]^(char)a[j];
        }
        for(i=0;i<7;i++) cout<<(char)s[i];
        cout<<"\n";
    }
}

Find the Lost Sock