It is said that 90% of frosh expect to be above average in their class. You are to provide a reality check.

The first line of standard input contains an integer C, the number of test cases. C data sets follow. Each data set begins with an integer, N, the number of people in the class (1 <= N <= 1000). N integers follow, separated by spaces or newlines, each giving the final grade (an integer between 0 and 100) of a student in the class.

For each case you are to output a line giving the percentage of students whose grade is above average, rounded to 3 decimal places.

5

5 50 50 70 80 100

7 100 95 90 80 70 60 50

3 70 90 80

3 70 90 81

9 100 99 98 97 96 95 94 93 91

40.000%

57.143%

33.333%

66.667%

55.556%

简单题 求分数超过平均分的学生的比例
来看一下WA和AC的代码
下面是WA的

 #include<stdio.h>

 int sco[+];

 int main()

 {

     int T,n,i;

     double ave;

     double res;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d",&n);

         ave=0.0;res=0.0;

         for(i=;i<n;i++)

         {

             scanf("%d",&sco[i]);

             ave+=double(sco[i]);

         }

         ave/=double(n);

         for( i=;i<n;i++)

         {

             if(sco[i]>ave)res+=1.0;

         }

         res/=n;res*=;

         printf("%.3lf%%\n",res);

     }

     return ;

 }

 #include<stdio.h>

 int sco[+];

 int main()

 {

     int T,n,i;

     float ave;

     float res;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d",&n);

         ave=0.0;res=0.0;

         for(i=;i<n;i++)

         {

             scanf("%d",&sco[i]);

             ave+=float(sco[i]);

         }

         ave/=float(n);

         for( i=;i<n;i++)

         {

             if(sco[i]>ave)res+=1.0;

         }

         res/=n;res*=100.0;

         printf("%.3f%%\n",res);

     }

     return ;

 }