Find a number HNUST OJ 1517 (位运算 判断一列数字中出现的唯一一个奇数数字)

题目描述

Find a number which is repeated odd times, then You should output the number.

Example 1:

if input is:12 12 12 12 15

then output is: 15

Example 2:

if input is:12 13 12 13 18 12 13 13 18

then output is: 12

输入

First line contains a positive integer N < 500000 ,then, N positive integers follow (delimited with space) each less than 1 000 000.

输出

In input sequence only one number X is repeated odd times. Others have even number of occurrences. You should output X.

样例输入

9

3 1 2 2 17 1 3 17 3

样例输出

3

提示

If you can avoid error "Memory Limit Exceed", this problem will be a very simple problem.

思路：位运算中的异或运算符，2个数字异或则为0；

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
int n;
while(scanf("%d",&n)==1)
{
int sum=0,num;
while(n--)
{
scanf("%d",&num);
sum=sum^num;
}
printf("%d\n",sum);
}
}