Ignatius and the Princess IV

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面对一组包含特殊整数的序列，该整数至少出现（n+1）/2次，任务是找出并返回这个特殊整数。本文提供两种解决策略：一是通过排序找到位于序列中间的元素，二是使用哈希表记录每个元素的出现频率，找出出现次数最多的元素。

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Problem Description

"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.

Sample Input

1 3 2 3 3

1 1 1 1 1 5 5 5 5 5 5

1 1 1 1 1 1 1

Sample Output

意思就是：一个整数至少出现（n + 1）/ 2次，咱们就是找出来这个数，

有以下做法：

1、/*刚好这个次数比较特殊，为数列的中间
所以，可以先进行排序，
那么寻找的这个数肯定有一半的出现次数，
即排序后，一半位置都是这个数*/
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int a[999999];
int main()
{
   int n;
   while(~scanf("%d",&n))
   {
       for(int i=0;i<n;i++)
       cin>>a[i];
       sort(a,a+n);
       printf("%d\n",a[(n+1)/2]);
   }
}

2、

/*每输入一个数，已此数为脚码，+1；最终进行比较，输出最大的数组，再输出其脚码*/

#include<cstdio>
#include<cstring>
#include<iostream>
#define N 1000000
int a[N];
using namespace std;
int main() {
   int s,i,m,k,t;
   while(~scanf("%d",&t)) {
       memset(a,0,sizeof(a));
       for(i=0; i<t; i++) {
           scanf("%d",&k);
           a[k]++;
       }
       m=0;
       for(i=0; i<1000000; i++) {
           if(a[i]>m) {
               m=a[i];
               s=i;
           }
       }
       printf("%d\n",s);

}
}