Distribution money（水题--模拟）

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本文详细解析了HDOJ上的竞赛题目「分布金钱」，通过直接模拟的方法解决了一个涉及金钱分配和惩罚条件的数学问题。该题旨在通过输入的人数和每个人ID的分配情况，找出可能受到惩罚的人员ID。

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5364

Distribution money

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 19 Accepted Submission(s): 13

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

Sample Output

  
   1
-1

Source

BestCoder Round #50 (div.2)

编程思想：按题意直接模拟。

AC code：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
#define ll __int64
#define LL long long 
#define MAXN 1000010
using namespace std;
const int INF=0x3f3f3f3f;
const int N = 100005;
const int mod = 1000000007;
LL a[MAXN];
int main()
{
    int t,n,m,mi,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        m=-INF;
        mi=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&t);
            a[t]++;
            if(a[t]>m)
            {
                m=a[t];
                mi=t;
            }
        }
        if(m>n-m)
        {
            printf("%d\n",mi);
        }
        else
        {
            printf("-1\n");
        }
    }
}