find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22110    Accepted Submission(s): 8746 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）

分析：一开始想用map，定义map<int,bool>类型，但是不好取key值，所以就用了set，用set也不太会遍历，然后现学了一下，发现，用迭代器就可以，这里我理解为，迭代器就是一个基类型的指针，然后输出里面的值只需要用*p即可，这个题最后只剩下一个值，所以直接用*ans.begin()就可以了。

代码如下：

#include <stdio.h> #include <set> using namespace std; set <int> ans; int main() { int n; int i,j; while(scanf("%d",&n),n) { int a; ans.clear(); for(i=0;i<n;i++) { scanf("%d",&a); if(ans.find(a)!=ans.end()) ans.erase(a); else ans.insert(a); } printf("%d\n",*ans.begin()); } return 0; }