hdu 2095 find your present (2) 找到只出现一次的数字

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 15349    Accepted Submission(s): 5821

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

数据有点水, 不符合要求的都是成对出现,所以有人用位运算^给A了  

我是用map做, 之前还想排序下搞,发现内存限制,,,, 1024*1024 / 4=262144   无法满足要存的10^6

所以,只能map了.

#include <stdio.h>#include <algorithm>#include <map>using namespace std;int main(){int n,i,a;map<int,int>my;map<int,int>::iterator it;while(scanf("%d",&n),n){for(i=0;i<n;i++) {scanf("%d",&a);my[a]++;}for(it=my.begin();it!=my.end();it++){if(it->second==1){printf("%d\n",it->first);break;}}my.clear();}return 0;}