HDU-1029 Ignatius and the Princess IV

Ignatius and the Princess IV

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

题意：给你一个奇数N，求N个书中出现次数超过(N+1)/2的数；

思路：排序，出现次数超过(N+1)/2的数一定在排序后的中间位置。

#include <iostream>

#include <cstdio>

#include <algorithm>

using namespace std;

#define maxn 1000000

int a[maxn];

int main()

{

    int n;

    while(~scanf("%d",&n))

    {

        for(int i=0;i<n;i++)

        {

            scanf("%d",&a[i]);

        }

        sort(a,a+n);

        printf("%d\n",a[(n+1)/2]);

    }

    return 0;

}

还有一种方法就是计数。（以下代码来源于网络）

1. #include <stdio.h>  

2. #include <string.h>  

3. #define MAX 5000010  

4.   

5. int dp[MAX] ;  

6.   

7. int main()  

8. {  

9.     int n ;  

10.     while(scanf("%d",&n)!=EOF)  

11.     {  

12.         int index = 0 ;  

13.         memset(dp,0,sizeof(dp)) ;  

14.         for(int i = 0 ; i < n ; ++i)  

15.         {  

16.             int t;  

17.             scanf("%d",&t) ;  

18.             dp[t]++ ;  

19.             if(dp[t]>=(n+1)/2) index = t;  

20.         }  

21.         printf("%d\n",index) ;  

22.     }  

23.     return 0 ;  

24. }