PAT 1005 Spell It Right [字符串处理]

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

Input Specification:

Each input file contains one test case. Each case occupies one line which contains an N (≤10​100​​).

Output Specification:

For each test case, output in one line the digits of the sum in English words. There must be one space between two consecutive words, but no extra space at the end of a line.

Sample Input:

12345

Sample Output:

one five

我的代码： 

//非递归
 
#include<stdio.h>
#include<string.h>

int main()
{
	char num[10][10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
	char s[100];
	scanf("%s",s);
	int i,sum = 0,t[100],count = 0;
	for(i=0;i<strlen(s);i++)
		sum += s[i]-'0';
	//特殊情况千万不要漏！（如果输入为0，就进不去下面的while循环 
	if(sum==0)
	{
		printf("zero");
		return 0;
	}
	while(sum!=0)
	{
		t[count++] = sum%10;
 		sum /= 10;
	}
	//逆序输出 
	for(i=count-1;i>=0;i--)
	{
		printf("%s",num[t[i]]);
		if(i) printf(" ");
	}       
}

//递归
#include<stdio.h>
#include<string.h>

char num[10][10]={"zero","one","two","three","four","five","six","seven","eight","nine"};

//递归不需要考虑zero的特殊情况，也不需要逆序输出（栈的思想）
void f(int n)
{
	if(n/10==0)
	{
		printf("%s",num[n%10]);
		return;
	}
	f(n/10);
	printf(" %s",num[n%10]);
} 

int main()
{
	char s[100];
	scanf("%s",s);
	int i,sum = 0,t[100],count = 0;
	for(i=0;i<strlen(s);i++)
		sum += s[i]-'0';  
	f(sum);
}