10370 - Above Average

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.

据说，90%的大学一年级新生期望他们自己的成绩能在全班平均之上，请你来帮忙验证看看他们有没有达成目标。

Input

输入的第一列有一个整数C 代表以下有多少组测试资料。每组资料的第一个整数N 代表班级总人数( 1 <= N <= 1000 )。接下来有N个以空白或换行来间隔的整数，代表每个学生的期末总成绩( 0 <= 分数<= 100 )。
　

Output

对每组测试资料输出一列，算出有多少百分比的学生成绩比全班平均高，请四舍五入到小数第三位。

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

Output for Sample Input

40.000%
57.143%
33.333%
66.667%
55.556%

解题思路：按照题目给定的要求，先求平均数再算比例，这体制要个小数有关的就定义为double

#include<stdio.h>
int main()
{int n,s,a[1000]={0},i,b;
double sum,t;
scanf("%d",&n);
while(n--){t=0;
           scanf("%d",&s);
           for(sum=i=0;i<s;i++){
                            scanf("%d",&a[i]);
                            sum+=a[i];
                            }
           sum/=s;
           for(t=i=0;i<s;i++)
           if(a[i]>sum)t++;
           printf("%.3f%%\n",t/s*100);
           }
return 0;
}