编程实现:任意输入一个英文句子,直到输入“.”结束,统计单词个数、空格个数并输出；这个程序怎么编呀

#include<stdio.h>
int main(void)
{
    char c;
    int a=0,b=0;
    printf("Input string:");
    while((c=getchar())!='.')
    {
        if(c>='a'&&c<='z'||c>='A'&&c<='Z')
            a++;
        if(c==' ')
            b++;
    }
    printf("句子中的空格数是：%d\n",b);
    printf("句子中的字母数是：%d\n",a);
    return 0;
}

再加个容错吧  else
                c++;
             printf("句子中的其他字符数是：%d\n",c);

谢谢啦，研究研究！！！

2楼正解

#include<stdio.h>
#include<string.h>
#define N 200
void main()
{
   int ch=0,num=0,other=0,i=0;
   char a[N];
   gets(a);
   for(i=0;i<=N-1;i++)
   {  
       if(a[i]!='\0')
       {
           if((a[i]>='A' && a[i]<='Z') || (a[i]>='a' && a[i]<='z'))\
             {ch++;}
    else if(a[i]>='0' && a[i]<='9')
       {num++;}
    else other++;
       }
       else break;
   }
   printf("ch:%d\nnum:%d\nother:%d",ch,num,other);
}

#include<stdio.h>
main()
{
    int c,flag=1,space=0,word=0;
    while((c=getchar())!='.')
    {
   
         if(c==' ')
         {
        space++;
        flag=1;
         }
   
        else if(flag==1)
        {
        word++;
        flag=0;   
        }
        
    }
    printf("%d\n%d",space,word);
}

#include <stdio.h>
void main( )
{
    char ch;
   
            int en=0,digital=0,space=0,other=0;
   
           while((ch=getchar()) != '\n')
    {
        if((ch>='a' && ch<='z') || (ch>='A' && ch<='Z'))
        {   
            en++;
        }
        else
            if(ch>='0' && ch<='9')
            {   
                digital++;
            }
            else
                if(ch==' ')
            {
                    space++;
}
            else
            {   
                other++;
            }
}
    printf("字母 %d 个\n",en);
   
            printf("数字 %d 个\n",digital);
   
            printf("空格 %d 个\n",space);
   
            printf("其他 %d 个\n",other);
}