HDU 2029

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#HDU #基础题

本文介绍了一个算法问题：在一组由奇数个整数组成的数据中找出至少出现(N+1)/2次的特殊整数。通过使用C++中的map实现散列表进行计数，有效地解决了这一挑战。

/*
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

*/
//题意给定一个奇数N,然后输入N个整数，统计其中出现（N+1/2）次数的数。
//一想到记数，果断散列表上去了！第一次这么如意地一次A过啊！

#include <iostream>
#include <cstdio>
#include<map>

using namespace std;

int main()
{
    int N,count1,t;
    map<int,int>get;
    while(scanf("%d",&N)!=-1){
            get.clear();//清空散列表
            count1=(N+1)/2;
        for(int i=0;i<N;i++){
            scanf("%d",&t);
            get[t]++;
        }
        map<int,int>::iterator it;
        for(it=get.begin();it!=get.end();it++){
            if(it->second>=count1)
                printf("%d\n",it->first);
        }
    }
    return 0;
}