Ignatius and the Princess IV（基础DP）

　"OK, you are not too bad, em... But you can never pass the next test." feng5166 says. 

　"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says. 

　"But what is the characteristic of the special integer?" Ignatius asks. 

　"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says. 

　Can you find the special integer for Ignatius? 

 

Input

　The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file. 

Output

　For each test case, you have to output only one line which contains the special number you have found. 

题意：

在n个数里有一个数的个数至少是（N+1）/2，求这个数

思路：

第一种可以排序，取中间那个数，就是答案，这种方法很久之前在《啊哈算法》的某个小角落看到过

第二种动态规划，每读一个数都判断与之前结果是否相同，相同num++，否则--，当num减完了，更新当前结果

代码：（超短）

#include <stdio.h>

int main()
{
    int n, i, x, num, re;
    while(~scanf("%d", &n))
    {
        num = 0;
        for(i=0; i<n; i++)
        {
            scanf("%d", &x);
            if(num==0) re = x;
            if(x==re) num++;
            else num--;
        }
        printf("%d
", re);
    }
    return 0;
}