哎  写的都不错 已经最优化了  

#include<stdio.h>
main()
{
    int i;
    int sum=0,j=0,ave=0;
    for(i=1;i<100;i+=2)
    {
        sum+=i;
        j++;
    }
    ave=sum/j;
    printf("the ave is %d\n",ave);
}

#include"stdio.h"
main()
{
    int i,sum=0,av,num=0;
    for(i=1;i<100;i+2)
      {sum+=i;
       num++;
      }
    av=sum/num;
    printf("%d\n",av);
}
//哈哈

#include <stdio.h>
int main(void)
{
    int i,sum=0,n=0;
    for(i=1;i<=100;i++){
        if(i%2==0)
            sum=sum;
        else
            sum=sum+i,n++;
    }
        printf("The average is:%d",sum/n);
        getchar();
        return 0;
}

#include "stdio.h"
void main()
{
   int i,sum=0;
   for(i=1;i<=100;i+=2)
      sum+=i;
   printf("1到100之间的奇数的和为%d\n",sum);
}

楼主的是太过繁琐了.....

只要求平均数而以，不一定要求和。。。
#include<stdio.h>
/*for the average of odd number between n to m */
int main()
{
    int n,m,ave;
    while(scanf("%d %d",&n,&m)!=EOF)   //  n<=m
    {
        ave=oddave(n,m);
        printf("%d\n",ave);
    }
}

int oddave(int n,int m)
{
    if((n%2 == 0) && (m%2 == 1))
    {
        return (n+1+m)/2;
    }
    else if((n%2 == 1) && (m%2 == 0))
    {
        return (n+m-1)/2;
    }
    else    //(((n%2 == 1) && (m%2 == 1))||((n%2 == 0) && (m%2 == 0)))
    {
        return (n+m)/2;
    }
}

#include <stdio.h>
void main(void)
{
    int i;
    int sum=0;
    int q=0;
    int p=0;

   
    for(i=1;i<=99;++i)
    {
      if(i%2 == 1)
      {
          q=q+i;
          sum=sum+1;
         
      }

    }
   
     p=q/sum;
   
    printf("p=%d\n",p);
    printf("%d\n",q);
    printf("%d\n",sum);
}
我这么写的

//求1到100之间奇数和的平均值
# include <stdio.h>
int main(void)
{
    int i, sum = 0, Num = 0, j;
    for (i=1; i<=100; i+=2)
    {
        sum = sum + i;      //奇数和
        Num += 1;          //奇数个数
    }
    j=sum/Num;
    printf("平均值 j = %d\n", j);
    return 0;
}
我这种方法行吗？

[ 本帖最后由 huang1982 于 2012-2-13 23:59 编辑 ]

for循环得用{}括起来  就和主函数一样