#include
int main(void)
{
int n,word,i;
char a;
n=0;
word=1;
while((a=getchar())!='&')
{
if(a>'a'&&a<'z')
n++;
if(a>'A'&&a<'Z')
n++;
if(a=' ')
word++;
}
i=n/word;
printf("单词平均有%d个字母\n",i);
}
结果为0;
做除法时转化为double,如i=n*1.0/word;把i定义为double
把i用双精度定义之后,把i=n/word改为i=(double)n/(double)word也行,强制类型转换挺好用