#include<stdio.h>
#include<ctype.h>
void main()
{
char answer='N';
double total=0.0;
double value=0.0;
int count=0;
printf("this program caculated the average of");
for( ;; )
{
printf("\nenter a value:");
scanf("%f",&value);
total+=value;
++count;
printf("do you want to enter another value？( N or Y ):");
scanf("%C",&answer);
if(tolower(answer)=='n')
break;
}
printf("\nthe average iS %.2f\n",total/count);
system("pause");
}
这个程序不能求平均值，为什么呢？

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

int main(void)
{
    int count = 0;
    double value = 0.0, total = 0.0;
    printf("this program caculated the average of:");
    for (count = 0; scanf("%lf", &value) == 1; count++)
        total += value;
    printf("the average iS %.2f\n", total / count);
    system("pause");
    return 0;
}

测试：
this program caculated the average of:12 13 14 end
the average iS 13.00
请按任意键继续. . .

[[it] 本帖最后由 风居住的街道 于 2008-10-27 22:35 编辑 [/it]]

tolower(answer)//system("pause");啥意思？

LS：一个问题，%C接受wchar_t而不是char，这样会导致内存泄露。
如果非要这么写，考虑这个实现（注意%c前面的空格）：
scanf(" %c", &answer);