BestCoder Round #50 (div.2)--------A Distribution money

Distribution money

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

AFA want to distribution her money to somebody.She divide her money into n same parts.One who want to get the money can get more than one part.But if one man's money is more than the sum of all others'.He shoule be punished.Each one who get a part of money would write down his ID on that part.

Input

There are multiply cases. For each case,there is a single integer n(1<=n<=1000) in first line. In second line,there are n integer a1,a2...an(0<=ai<10000)ai is the the ith man's ID.

Output

Output ID of the man who should be punished. If nobody should be punished,output -1.

Sample Input

3
1 1 2
4
2 1 4 3

Sample Output

1
-1

分析：这是我第三次打BC，还不是很习惯。这道题很水，我却想多了，代码写得乱。

思路：每份钱一样，所以同一个人领取的份数最多的超过总份数的一半就惩罚他。注意：当同一个人领取时也会受到惩罚。

CODE：

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

int main()
{
    int arr[10005],a[10005],brr[10005];
    int n;
    while(cin>>n){
        memset(arr,0,sizeof(arr));
        memset(brr,0,sizeof(brr));
        for(int i=0;i<n;i++){
            cin>>a[i];
            arr[a[i]]++;
        }
        if(n==1){
            cout<<-1<<endl;
            continue;
        }
        if(arr[a[0]]==n){
            cout<<a[0]<<endl;
            continue;
        }
        for(int i=0;i<10005;i++)
            brr[i]=arr[i];
        sort(brr,brr+10005);
        if(brr[10004]!=brr[10003]&&brr[10004]*2>n){
            for(int i=0;i<n;i++){
                if(arr[a[i]]==brr[10004]){
                    cout<<a[i]<<endl;
                    break;
                }
            }
        }
        else
            cout<<-1<<endl;
    }
}