HDU 1029 Ignatius and the Princess IV （动态规划、思维）

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 51503    Accepted Submission(s): 23178

 

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

5
1 3 2 3 3
11
1 1 1 1 1 5 5 5 5 5 5
7
1 1 1 1 1 1 1

Sample Output

3
5
1

题目大意与思路

输入一组数，n个，将该组数中相同数字的个数大于（n+1）/ 2 的数字输出。

我是看题目分类来做这个题的，怎么也想不出来动规怎么做 看了别人的题解恍然大悟 真是太高妙了

如果这个数出现次数大于（n+1）/ 2的话 那么他比所有其他的数出现次数都要多！

具体的看代码吧 应该很清楚了

#include<bits/stdc++.h>

using namespace std;

int i,n,cnt,anss,x;

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        cnt=0;
        for(i=1;i<=n;i++)
        {
            cin>>x;
            if(cnt==0)
            {
                cnt++;
                anss=x;
            }
            else
            {
                if(x==anss)
                cnt++;
                else
                cnt--;
            }        
        }
        cout<<anss<<endl;
    }
}