HDU2095(异或运算)

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2095

解题思路：

把所有数都做异或运算，结果一定是出现奇数的那个数。因为出现偶数次的数，做异或后为0.

完整代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;int main(){    #ifdef DoubleQ    freopen("in.txt","r",stdin);    #endif    int n;    while(~scanf("%d",&n))    {        if(n == 0)            break;        int k;        int sum;        for(int i = 0 ; i < n ; i ++)        {            scanf("%d",&k);            if(i == 0)                sum = k;            else                sum ^= k;        }        printf("%d\n",sum);    }}

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