这个题目明显用数组比用switch简单

#include<stdio.h>
void main()
{
char a[10][10]={"zero","one","tow","tree","four","five","six","seven","eght","nine"};
int i;
char b[10];
while(scanf("%s",b)!=EOF)
{
for(i=0;b[i];i++)
printf("%s ",a[b[i]-'0']);
printf("\n");
}
}

用指针，或者数组，加个switch就可以了，呵呵好心都搞定了啊。

crackerwang
一直拉下来看,终于看到一个是这样做的.

我改了以下你的代码
#include<stdio.h>
void main()
{
char a[10][10]={"zero","one","tow","tree","four","five","six","seven","eght","nine"};
int i;
while(scanf("%1d",&i)!=EOF)
{
printf("%s ",a[i]);
}
}

mp3改的好简洁，佩服

厉害.
我怎么就没有想到呢?
高手就是高手

#include<stdio.h>
void main()
{
char a[10][10]={"zero","one","tow","tree","four","five","six","seven","eght","nine"};
char c;
while((c=getchar())!='\0')
{
printf("%s ",a[c-'0']);
}
}

呵呵nuciewth斑竹 scanf("%1d",&amp;i) 我这里用的是 1d 也就是说每次有只有1位数存入i

斑竹看错了估计看成了ld了 long int
哈哈
ld
1d
谁能用眼睛分的清楚