HDU1029 Ignatius and the Princess IV

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 27552    Accepted Submission(s): 11728

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

51 3 2 3 3111 1 1 1 1 5 5 5 5 5 571 1 1 1 1 1 1

 

 

Sample Output

351

 

        这道题目求的是出现次数超过总次数一半的元素我们可以总结

发现一个规律，就是在一个多数元素为n的数列中，去掉两个不相同

的元素，剩下数列的多数元素还是n进一步推广，去掉的元素中的多

数元素如果不是n，那么剩下数列的多数元素任然是n，这个可以通

过一个出现频率的概念来理解。假设n的出现频率为0.5，而前半段

的出现频率小于0.5那么后半段的出现频率必然大于0.5。

        想清楚这点以后这道题目就可以解决了，值得注意的是没有必要

通过数组来存储每一个数据，直接在每次输入数据的时候进行操

作，可以节省内存空间的使用。

 

代码：

#include <iostream>#include <cstdio>using namespace std;int main(){    int n,t,res;    while(~scanf("%d",&n))    {        int cnt=0;        for(int i=0;i<n;i++)        {            scanf("%d",&t);        if(cnt==0)        {            cnt=1;            res=t;        }        else        {            if(t==res) cnt++;            else            {                cnt--;            }        }        }        printf("%d\n",res);    }    return 0;}