C语言实现 计算句子的平均句长

#include<stdio.h>
int main()
{
   char ch,a[80]={'
 
 
#include<stdio.h>
int main()
{
   char ch,a[80]={'\0'};
   int i=0,n=0,m=0,j=1;
   printf("Enter a first and last name:");
   ch=getchar();
   while(ch!='\n')
   {
      a[i++]=ch;
      ch=getchar();
   }
   a[i]='\0';
   for(n=0;n<i;n++)
   {
     if(a[n]==' ')
     {
         j++;
         continue;
     }
      m++;

   }
   printf("Averge word length:%.1f",(float)m/j);
   getch();
}
 
 
   
  
  

'};
   int i=0,n=0,m=0,j=1;
   printf("Enter a first and last name:");
   ch=getchar();
   while(ch!='\n')
   {
      a[i++]=ch;
      ch=getchar();
   }
   a[i]='
 
 
#include<stdio.h>
int main()
{
   char ch,a[80]={'\0'};
   int i=0,n=0,m=0,j=1;
   printf("Enter a first and last name:");
   ch=getchar();
   while(ch!='\n')
   {
      a[i++]=ch;
      ch=getchar();
   }
   a[i]='\0';
   for(n=0;n<i;n++)
   {
     if(a[n]==' ')
     {
         j++;
         continue;
     }
      m++;

   }
   printf("Averge word length:%.1f",(float)m/j);
   getch();
}
 
 
   
  
  

';
   for(n=0;n<i;n++)
   {
     if(a[n]==' ')
     {
         j++;
         continue;
     }
      m++;

   }
   printf("Averge word length:%.1f",(float)m/j);
   getch();
}#include<stdio.h>
int main()
{
   char ch,a[80]