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
1 #include<stdio.h> 2 int sco[1000+10]; 3 int main() 4 { 5 int T,n,i; 6 double ave; 7 double res; 8 scanf("%d",&T); 9 while(T--) 10 { 11 scanf("%d",&n); 12 ave=0.0;res=0.0; 13 for(i=0;i<n;i++) 14 { 15 scanf("%d",&sco[i]); 16 ave+=double(sco[i]); 17 } 18 ave/=double(n); 19 for( i=0;i<n;i++) 20 { 21 if(sco[i]>ave)res+=1.0; 22 } 23 res/=n;res*=100; 24 printf("%.3lf%% ",res); 25 } 26 return 0; 27 }
下面是AC的
1 #include<stdio.h> 2 int sco[1000+10]; 3 int main() 4 { 5 int T,n,i; 6 float ave; 7 float res; 8 scanf("%d",&T); 9 while(T--) 10 { 11 scanf("%d",&n); 12 ave=0.0;res=0.0; 13 for(i=0;i<n;i++) 14 { 15 scanf("%d",&sco[i]); 16 ave+=float(sco[i]); 17 } 18 ave/=float(n); 19 for( i=0;i<n;i++) 20 { 21 if(sco[i]>ave)res+=1.0; 22 } 23 res/=n;res*=100.0; 24 printf("%.3f%% ",res); 25 } 26 return 0; 27 }
就是把double改成了float......