hdoj.2095 find your present (2) 20140804

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 15325    Accepted Submission(s): 5813

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

51 1 3 2 231 2 10

 

Sample Output

32
Hint
Hint
 use scanf to avoid Time Limit Exceeded
 
注：此题若采用常规方法会超内存，且题目大致意思为找出n个数中出现奇数次的数，可以采用C语言异或（^）运算符进行操作。某个数本身异或偶数次的结果是0,0与任何数的异或为任何数。
 
#include<stdio.h>int main(){    int n,i,x,y;    while(scanf("%d",&n)!=EOF&&n)    {           x=0;           while(n--)           {                scanf("%d",&y);                x^=y;                }                    printf("%d\n",x);                 }    return 0;}