BestCoder Round #50 (div.2) 1001

Distribution money

Accepts: 713

Submissions: 1881

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

31 1 242 1 4 3

Sample Output

1-1

签到题，当前序号下所占的份数大于0.5，则输出该序号，都没有则输出-1

默默地吐槽杭电不支持#include<bits/stdc++.h>

//#include<bits/stdc++.h>#include<cstdio>#include<iostream>using namespace std;int n;struct div{    double b;}ans[10100];int main(){    //freopen("in.txt","r",stdin);    while(~scanf("%d",&n))    {        int x,count=-1;        for(int i=0;i<n;i++)            ans[i].b=0;        for(int i=0;i<n;i++)        {            scanf("%d",&x);            ans[x].b+=1.0/n;            if(ans[x].b>(1.0/2))                count=x;        }        printf("%d\n",count);    }    return 0;}