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[導(dǎo)讀]題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=5364 題面: Distribution money Time Limit: 2000/1000 M

題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=5364


題面:

Distribution money Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 655????Accepted Submission(s): 343


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
3
1 1 2
4
2 1 4 3

?

Sample Output
1
-1

?
解題:

??? 計個數(shù)的事...

代碼:

#include 
#include 
#include 
using namespace std;
int cnt[10010];
int main()
{
	int n,tmp;
    while(~scanf("%d",&n))
	{
	   bool flag=true;
	   memset(cnt,0,sizeof(cnt));
	   for(int i=1;i<=n;i++)
	   {
		   scanf("%d",&tmp);
		   cnt[tmp]++;
		}
	   for(int i=1;i<=10000;i++)
	   {
		   if(cnt[i]>(n-cnt[i]))
		   {
			   printf("%dn",i);
			   flag=false;
			   break;
		   }
	   }
	   if(flag)printf("-1n");
	}
}


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