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Problem Description
假设一个班有n(n<=50)个学生，每人考m(m<=5)门课，求每个学生的平均成绩和每门课的平均成绩，并输出各科成绩均大于等于平均成绩的学生数量。
Input
输入数据有多个测试实例，每个测试实例的第一行包括两个整数n和m，分别表示学生数和课程数。然后是n行数据，每行包括m个整数（即：考试分数）。
Output
对于每个测试实例，输出3行数据，第一行包含n个数据，表示n个学生的平均成绩，结果保留两位小数；第二行包含m个数据，表示m门课的平均成绩，结果保留两位小数；第三行是一个整数，表示该班级中各科成绩均大于等于平均成绩的学生数量。
每个测试实例后面跟一个空行。
Sample Input
2 2
5 10
10 20
Sample Output
7.50 15.00
7.50 15.00
1
#include<stdio.h> int main() { long long int n,m,i,j,a[50][50],s1,k,q,s2,count,
p; double xp[50],kp[50]; while(~scanf("%lld%lld",&n,&m)) { count=0;k=0;q=0; for(
i=0;i<n;i++) { s1=0; for(j=0;j<m;j++) { scanf("%lld",&a[i][j]); s1+=a[i][j]; }
xp[k++]=s1*1.0/m; } for(j=0;j<m;j++) { s2=0; for(i=0;i<n;i++) { s2+=a[i][j]; }
kp[q++]=s2*1.0/n; } for(i=0;i<n;i++) { p=1;q=0; for(j=0;j<m;j++) { if(a[i][j]<kp
[q++]) {p=0;break;} } if(p) count++; } k=0; for(i=0;i<n-1;i++) printf("%.2lf ",
xp[k++]); printf("%.2lf\n",xp[k]); q=0; for(i=0;i<m-1;i++) printf("%.2lf ",kp[q
++]); printf("%.2lf\n",kp[q]); printf("%lld\n\n",count); } return 0; }
