编写一个程序使之读入一系列整型数据直到读入标志值-1时为止。此时程序显示出之前所输入数据的平均值，编写的程序应能产生下面的示例
this program averages a list of integers.
enter -1 to signal the end of the list.
?95
？100
？89
？97
？91
？97
？-1
the average is 94.4

#include<stdio.h>
int main(void)
{
 int i;
 char ch;
 float sum,a;
 i=0;
 sum=0;
 while(a!=-1)
 {
   ch=getchar();
   if(ch=='?')
 {
  i++;
  scanf("%f",&a);
  sum=sum+a;
 }
 }
 printf("the average is %f",sum/(i-1));
return 0;
}

#include<stdio.h>
void main()
{
    int b,p,n=1,sum=0;
    int avr=0;
    int a[100];
    while(1)
    {
        printf("?  ");
        scanf("%d",&p);
        if(p!=-1)
        {
            a[n]=p;
            n++;
        }
        else
            break;
    }
    for(b=1;b<n;b++)
    sum=sum+a[b];
    avr=sum/(n-1);
    printf("avr=%d\n",avr);
}




这样么？

[[it] 本帖最后由 小小熊宝宝 于 2008-9-20 21:34 编辑 [/it]]

不是这样，别人要小数显示，这样的程序空间浪费大,sum还要溢出的可能.

[[it] 本帖最后由 iFreeBSD 于 2008-9-20 21:29 编辑 [/it]]

LZ帮你写了有什么好处

楼上的写个出来我研究研究学习学习~呵呵

main()
{int num,sum=0,n=0;
float average;
printf("this program averages a list of integers.\n");
printf("enter -1 to signal the end of the list.\n");
printf("?");
scanf("%d",&num);
while(num!=-1)

{n=n+1;
sum=sum+num;
printf("?");
scanf("%d",&num);}
average=sum/n;
printf("%f\n",average);

}

#include <stdio.h>
int main(void) {
          float  avg = 0.0F ;
          int  i , current_integer ;
          char *msg = "this program averages a list of integers.\n"
                      "enter -1 to signal the end of the list.\n" ;
          printf(msg) ;
          for (i = 1 ;putchar('?') ,scanf("%d",&current_integer) && current_integer != -1;i++)
                  avg += (current_integer - avg) / i ;  
            
          printf("the average is %.1f\n" , avg) ;
 return 0 ;
}

#include<stdio.h>
void main()
{
    int b,p,n=1;
    float avr=0,sum=0;
    int a[100];
    while(1)
    {
        printf("?  ");
        scanf("%d",&p);
        if(p!=-1)
        {
            a[n]=p;
            n++;
        }
        else
            break;
    }
    for(b=1;b<n;b++)
    sum=sum+a[b];
    avr=(float)sum/(n-1);
    printf("avr=%4.2f\n",avr);
}

这个问号我还以为是输入的条件呢~理解错了~不好意思LZ看下面的吧