HDU 2095 find your present (2)

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 13413    Accepted Submission(s): 5049

Problem 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

 

Sample Output

3

2

Hint

Hint

use scanf to avoid Time Limit Exceeded

 

Author

8600

 

Source

HDU 2007-Spring Programming Contest - Warm Up （1）

 

Recommend

8600   |   We have carefully selected several similar problems for you:  2094 1597 1593 1595 1599 

 

思路：异或运算“^”，异或运算是（二进制）按位相同取0，按位不同取1，例如5 — 101，3 — 011,5 ^ 3 = 110 = 6，我、而且满足交换律，result ^ 2 ^ 1 ^ 1 ^ 2 = result ^ 2 ^ 2 ^ 1 ^ 1，而2 ^ 2 = 0，所以最后的结果是所有偶数次数的都为零，最后的结果就是所求

 

代码：

#include<iostream>
using namespace std;
int main()
{
    int num,n,result,i;
    while(scanf("%d",&n),n){
        result=0;
        for(i=0;i<n;++i){
            scanf("%d",&num);
            result^=num;
        }
        printf("%d\n",result);
    }
    return 0;
}