http://acm.hdu.edu.cn/showproblem.php?pid=536
Distribution money
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 345 Accepted Submission(s):
206
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.
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.
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.
If
nobody should be punished,output -1.
Sample Input
3
1 1 2
4
2 1 4 3
Sample Output
1
-1
Source
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a,h[];
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(h,,sizeof(h));
int i;
for(i=;i<n;i++)
{
scanf("%d",&a);
h[a]++;
}
int max=,t;
for(i=;i<=;i++)
{
// printf("h=%d\n",h[i]);
if(h[i]>=max)
{
max=h[i];
t=i;
}
}
if(max>(n-max))
printf("%d\n",t);
else
printf("-1\n"); }
return ;
}