Sicily 2005. Lovely Number

最新推荐文章于 2018-01-04 15:02:37 发布

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本文介绍了一种通过使用哈希映射（map）来找出数组中唯一出现奇数次的元素的方法。面对由偶数次重复元素组成的数据集，除了一个特定的‘可爱数字’以外，任务是在最短时间内定位这个唯一的奇数次元素。

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Time Limit: 1 secs, Memory Limit: 64 MB

Description

Every time after got a solution from kevin, ivan repeats again and again,”Are there any better ones?”. It seems more worse this time. Given a sequence of numbers, which ivan is sure that all of them appear even times except only one “lovely number” appears odd times, kevin must find out the lovely number as soon as possible. So, let’s rock!

Input

There are multiple test cases.

Each test case contains two lines. The first line is an integer N, 1 <= N <= 10001, and N % 2 = 1. The second line contains N integers, separated by spaces.

Input is terminated by EOF.

Output

For each test case, output a line of an integer, which should be the lovely number.

Sample Input

3
1 2 1
5
7 7 7 7 3
7
2 2 4 4 4 2 2

Sample Output

2
3
4
Problem Source

2010中山大学新手赛-网络预选赛 ivankevin@argo

(。＾▽＾) Just do it!

#include <iostream>
#include <map>

using namespace std;

int main()
{
    map<int, int> Temp;
    int T, Tp;
    while (cin >> T)
    {
        Temp.clear();
        while (T--) {
            cin >> Tp;
            Temp[Tp]++;
        }
        map<int, int>::iterator i = Temp.begin();
        while (i != Temp.end()) {
            if ((*i).second % 2 == 1) {
                cout << (*i).first << endl;
                break;
            }
            i++;
        }
    }
}