hdu 5364 Distribution money 简单题_hdu gusee the money-优快云博客

本文链接：https://blog.youkuaiyun.com/firenet1/article/details/47399503

本文探讨了如何通过算法公平地分配一定数额的资金，并在分配过程中识别并惩罚那些获得资金超过平均分配额的人。详细解释了算法的工作原理，包括输入案例的解析、输出规则的应用，以及如何使用代码实现这一过程。

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Distribution money

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 374 Accepted Submission(s): 224

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)

统计大于 n/2的数的个数

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
long long mod = 1000000007;

int num[10007];
int main(){
    int n,u;
    while(scanf("%d",&n)!=EOF){
        memset(num,0,sizeof(num));
        for(int i = 0;i < n; i++)
        {
            scanf("%d",&u);
            num[u]++;
        }
        int flag = -1;
        for(int i = 0;i < 10001; i++){
            if(num[i] > n-num[i])
                flag = i;
        }
        printf("%d\n",flag);
    }
    return 0;
}