Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103993    Accepted Submission(s): 24170
//#include<iostream>
//using namespace std;
#include<stdio.h> 
int main()
{
	int a[60][60];
	double  sum,ave1[60],ave2[60];
	int i,j,n,m,p,count;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i = 0;i < n;i++)
		{
			for(j = 0;j < m;j++)
			{
			      scanf("%d",&a[i][j]); 
			}
		}
		for(i = 0;i < n;i++)
		{
			sum = 0;
			for(j = 0;j < m;j++)
			{
				sum += a[i][j];
			}
			ave1[i] = sum/m;
		} 
		for(j = 0;j < m;j++)
		{
			sum = 0;
			for(i = 0;i < n;i++)
			{
				sum += a[i][j];
			}
			ave2[j] = sum/n;
		} 
		count = 0;
		for(i = 0;i < n;i++)
		{
			p = 1;
			for(j = 0;j < m;j++)
			{
				if(a[i][j] < ave2[j])
				{
					p = 0;
					break;
				}
			}
			if(p)
			{
				count++;
			} 
		} 
		for(i = 0;i < n-1;i++)
		{
			printf("%.2lf ",ave1[i]);
		}
		printf("%.2lf\n",ave1[i]);
		for(j = 0;j < m-1;j++)
		{
			printf("%.2lf ",ave2[j]);
		}
		printf("%.2lf\n",ave2[j]);
        printf("%d\n\n",count);
	}
	return 0;
} 
原文:http://www.cnblogs.com/lsh123/p/5866009.html