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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 74055 Accepted Submission(s):
17809
#include<stdio.h> #include<math.h> double n,m,a[70][7],k[9],y=0,u=0,sum=0; int i,j; int main() { while(scanf("%lf%lf",&n,&m)!=EOF) { for(i=1;i<=n;i++) { for(j=1;j<=m;j++) scanf("%lf",&a[i][j]); } for(i=1;i<=n;i++) // n个学生的平均成绩 { for(j=1;j<=m;j++) sum+=a[i][j]; if(i!=n) printf("%.2f ",sum/m); if(i==n) printf("%.2f\n",sum/m); sum=0; } sum=0; for(i=1;i<=m;i++) //m门课的平均成绩 { for(j=1;j<=n;j++) sum+=a[j][i]; k[i]=sum/n; if(i!=m) printf("%.2f ",k[i]); if(i==m) printf("%.2f\n",k[i]); sum=0; } for(i=1;i<=n;i++) { for(j=1;j<=m;j++) //班级中各科成绩均大于等于平均成绩的学生数量 { if(a[i][j]>=k[j]) y++; if(y==m) u++; } y=0; } printf("%.0f\n",u); printf("\n"); u=0; y=0; sum=0; } return 0; }
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原文地址:http://www.cnblogs.com/tonghao/p/4373418.html