Description
It is said that 90% of frosh expect to be above average in their class. You are to provide a reality check.
Input
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.
Output
For each case you are to output a line giving the percentage of students whose grade is above average, rounded to 3 decimal places.
Sample Input
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
Sample Output
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 ;
}
下面是AC的
#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 ;
}
就是把double改成了float......