7-25 read numbers (loop switch)

Enter an integer , Output the Pinyin corresponding to each number . When the integer is negative , First, the output fu word . The Pinyin corresponding to ten numbers is as follows ：

0: ling
1: yi
2: er
3: san
4: si
5: wu
6: liu
7: qi
8: ba
9: jiu

Input format ：
Input gives an integer on a line , Such as ：1234.

Tips ： Integers include negative numbers 、 Zero and positive .

Output format ：
Output the Pinyin corresponding to the integer in one line , The Pinyin of each number is separated by a space , There is no final space at the end of the line . Such as yi er san si.

sample input ：
-600
No blank lines at the end

sample output ：
fu liu ling ling
No blank lines at the end

There are two ways to achieve this ：

Code 1:

#include<stdio.h>
int main()
{
    int num = 0;
    int a[11] = { 0 };
    scanf("%d", &num);
    if (num < 0)
    {
        printf("fu ");
        num *= -1;
    }
    if (num == 0)
    {
        printf("ling");
    }
    int k = 0,i=0;
    while (num)// Store each bit of data in the array .
    {
        a[i] = num % 10;
        i++;
        num /= 10;
   }
    k = --i;
  for(int i=k;i>=0;i--)// Print each bit in turn , Last, don't print spaces 
  {
        switch (a[i])
        {
        case 0:
            if (i == 0)
            {
                printf("ling");
                break;
            }        
            printf("ling ");
            break;
        case 1:
            if (i == 0)
            {
                printf("yi");
                break;
            }            
            printf("yi ");
            break;
        case 2:
            if (i == 0)
            {
                printf("er");
                break;
            }              
            printf("er ");
            break;
        case 3:
            if (i == 0)
            {
                printf("san");
                break;
            }             
            printf("san ");
            break;
        case 4:
            if (i == 0)
            {
                printf("si");
                break;
            }              
            printf("si ");
            break;
        case 5:
            if (i == 0)
            {
                printf("wu");
                break;
            }                
            printf("wu ");
            break;
        case 6:
            if (i == 0)
            {
                printf("liu");
                break;
            }             
            printf("liu ");
            break;
        case 7:
            if (i == 0)
            {
                printf("qi");
                break;
            }              
            printf("qi ");
            break;
        case 8:
            if (i == 0)
            {
                printf("ba");
                break;
            }              
            printf("ba ");
            break;
        case 9:
            if (i == 0)
            {
                printf("jiu");
                break;
            }           
            printf("jiu ");
            break;
        }
    }
  return 0;
}

result ：

test result ：

Code 2：

 

#include<stdio.h>
 
int main()
{
	int num = 0;
	scanf("%d", &num);
	if (num < 0)
    {
        printf("fu ");
        num *= -1;
    }
		
	if (num == 0)
		printf("ling");
	int t = num;
	int i = 1;
	while (num)// This part is about num Figure out the number of digits 
	{
		i *= 10;
		num /= 10;
	}
	i = i / 10;
	int m = 0,d=0;
	while(i)// Print directly from the top 
	{
		m = t / i;
		d = m % 10;
		i /= 10;
		switch (d)
		{
		case 0: 
			if (i == 0)
			{
				printf("ling");
				break;
			}
			printf("ling "); break;
		case 1: 
			if (i == 0)
			{
				printf("yi");
				break;
			}
			printf("yi "); break;
		case 2:
			if (i == 0)
			{
				printf("er");
				break;
			}
			printf("er "); break;
		case 3: 
			if (i == 0)
			{
				printf("san");
				break;
			}
			printf("san "); break;
		case 4: 
			if (i == 0)
			{
				printf("si");
				break;
			}
			printf("si "); break;
		case 5: 
			if (i == 0)
			{
				printf("wu");
				break;
			}
			printf("wu "); break;
		case 6: 
			if (i == 0)
			{
				printf("liu");
				break;
			}
			printf("liu "); break;
		case 7: 
			if (i == 0)
			{
				printf("qi");
				break;
			}
			printf("qi "); break;
		case 8: 
			if (i == 0)
			{
				printf("ba");
				break;
			}
			printf("ba "); break;
		case 9: 
			if (i == 0)
			{
				printf("jiu");
				break;
			}
			printf("jiu "); break;
		}
	}
	return 0;
}

test result ：

  The problem itself is not difficult , It's easy to grasp the printing order of this problem , In fact, I have learned data structure , Stack to achieve the best .