Question

C语言代码求助。

C语言代码求助。已知10名同学的成绩（a[10]=80,70,60,50,20,80,9060,50,70）已存入数组a中，计算前n位同学的平均成绩，存入aver中，并输出aver。

Answer 1

#include "stdafx.h"
//#include "stdlib.h"
int main()
{
int a[10] = { 80, 70, 60, 50, 20, 80, 90, 60, 50, 70 };
int aver = 0;
for (int i = 0; i < 10; i++)
aver = aver + a[i];
aver = aver / 10;
printf("%d\n", aver);
//若闪退
//system("pause");
return 0;
}

追问

不对呀

追答

#include "stdafx.h"
//#include "stdlib.h"
int main()
{
int a[10] = { 80, 70, 60, 50, 20, 80, 90, 60, 50, 70 };
int aver = 0;
int n;
scamf("%d", &n);
for (int i = 0; i < n; i++)
aver = aver + a[i];
aver = aver / n;
printf("%d\n", aver);
//若闪退
//system("pause");
return 0;
}

追问

结果不对

超过10万了

追答

#include "stdafx.h"
//#include "stdlib.h"
int main()
{
double a[10] = { 80, 70, 60, 50, 20, 80, 90, 60, 50, 70 };
double aver = 0.0;
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
aver = aver + a[i];
aver = aver / n;
printf("%.2lf\n", aver);
//若闪退
//system("pause");
return 0;
}

追问

嗯。。。我晚上再试试

追答

还有这是保留两位的浮点数输出，要整数的话是这个：
#include "stdafx.h"
//#include "stdlib.h"
int main()
{
int a[10] = { 80, 70, 60, 50, 20, 80, 90, 60, 50, 70 };
int aver = 0;
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
aver = aver + a[i];
aver = aver / n;
printf("%d\n", aver);
//若闪退
//system("pause");
return 0;
}
还有//后面的代码没用到的化，所有有//在的行都可删掉

Answer 2

#include"stdio.h"
int main(){

int a[10]={80,70,60,50,20,80,90,60,50,70};
int aver=0;
int n;
scanf("%d",&n);
for(int i=0;i<n;i++){
aver+=a[i];
}
aver/=n;
printf("%d",aver);
return 0;
}