HDU 2095 find your present 位异或

Description

In the new year party, everybody will get a “special present”.Now it’s your turn to get your special present, a lot of presents now putting on the desk, and only one of them will be yours.Each present has a card number on it, and your present’s card number will be the one that different from all the others, and you can assume that only one number appear odd times.For example, there are 5 present, and their card numbers are 1, 2, 3, 2, 1.so your present will be the one with the card number of 3, because 3 is the number that different from all the others.

Input

The input file will consist of several cases.
Each case will be presented by an integer n (1<=n<1000000, and n is odd) at first. Following that, n positive integers will be given in a line, all integers will smaller than 2^31. These numbers indicate the card numbers of the presents.n = 0 ends the input.

Output

For each case, output an integer in a line, which is the card number of your present.

Sample Input

5
1 1 3 2 2
3
1 2 1
0

Smple Output

3
2
一般想到的是开数组，可是数据量太大了…
不过可以用异或处理,对输入的所有数进行异或，最后得到的就是答案，出现偶数次的数将会抵消

代码

#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
#include<vector>
#include<stdio.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        ll ans=0,x;
        for(int i=0;i<n;++i)
        {
            scanf("%lld",&x);
            ans=ans^x;
        }
        printf("%lld\n",ans);
    }
}