Distribution money（水题--模拟）

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

31 1 242 1 4 3

 

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 1000010using 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");        }    }}