HDU-1029 Ignatius and the Princess IV

最新推荐文章于 2021-10-03 22:24:43 发布

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最新推荐文章于 2021-10-03 22:24:43 发布

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面对feng5166的挑战，Ignatius必须从给定的奇数长度数列中找到那个至少出现(n+1)/2次的特殊整数，否则公主将面临危险。本文介绍了一种简单直接的方法来解决这个问题。

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Ignatius and the Princess IV

"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：给一个奇数n，长度为n的数列，找出其中出现次数不少于（n+1）/2的数字

思路：暴力

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
using namespace std;
map<int,int>mp;
int vis[1000005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int x,tot=1;
        mp.clear();
        memset(vis,0,sizeof(vis));
        int maxx=0,k;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            if(!mp[x]) mp[x]=tot++;
            vis[mp[x]]++;
            if(vis[mp[x]]>maxx)
                maxx=vis[mp[x]],k=x;
        }
        printf("%d\n",k);
    }
    return 0;
}