写出这个数----求指导系列

问题：

读入一个正整数 n，计算其各位数字之和，用汉语拼音写出和的每一位数字。

输入格式：

每个测试输入包含 1 个测试用例，即给出自然数 n 的值。这里保证 n 小于 10​100​​。

输出格式：

在一行内输出 n 的各位数字之和的每一位，拼音数字间有 1 空格，但一行中最后一个拼音数字后没有空格。

输入样例：

1234567890987654321123456789

输出样例：

yi san wu

小生不才： 

#include <iostream>
#include<string>
using namespace std;
int main()
{
	string a1;
	int i,sum,count=0;
	int a=0,b=0,c=0,d=0;
	int a2[4];
	string p[10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
	cin>>a1;
    int num = a1.length(); 
    for (int i = 0; i < num; ++i) 
    { 
         sum+=(int)a1[i]-48;
    }
    if(sum>=0)
    {
    	 a=sum/1000;
      	 b=(sum-1000*a)/100;
	     c=(sum-1000*a-100*b)/10;
	     d=(sum-1000*a-100*b-10*c);
	}

	if(0<a&&a<10)
	{
		switch(a)
		     {
		     	
		     	case 1:cout<<p[1]<<" ";break;
				 case 2:cout<<p[2]<<" ";break;
				 case 3:cout<<p[3]<<" ";break;
				 case 4:cout<<p[4]<<" ";break;
				 case 5:cout<<p[5]<<" ";break;
				 case 6:cout<<p[6]<<" ";break;
				 case 7:cout<<p[7]<<" ";break;
				 case 8:cout<<p[8]<<" ";break;
				 case 9:cout<<p[9]<<" ";break;
				 default:break;
			 }
	}
	if(a==0&&b==0)
	{
		switch(b)
		{
			default:break;
		}
	
	}
	else
	{
			switch(b)
		     {
		     	case 0:cout<<p[0]<<" ";break;
		     	case 1:cout<<p[1]<<" ";break;
				 case 2:cout<<p[2]<<" ";break;
				 case 3:cout<<p[3]<<" ";break;
				 case 4:cout<<p[4]<<" ";break;
				 case 5:cout<<p[5]<<" ";break;
				 case 6:cout<<p[6]<<" ";break;
				 case 7:cout<<p[7]<<" ";break;
				 case 8:cout<<p[8]<<" ";break;
				 case 9:cout<<p[9]<<" ";break;
				 default:break;
			 }
	}
	if(a==0&&b==0&&c==0)
	{
		switch(b)
		{
			default:break;
		}
	}
	else
	{
		switch(c)
		     {
		         case 0:cout<<p[0]<<" ";break;
		     	 case 1:cout<<p[1]<<" ";break;
				 case 2:cout<<p[2]<<" ";break;
				 case 3:cout<<p[3]<<" ";break;
				 case 4:cout<<p[4]<<" ";break;
				 case 5:cout<<p[5]<<" ";break;
				 case 6:cout<<p[6]<<" ";break;
				 case 7:cout<<p[7]<<" ";break;
				 case 8:cout<<p[8]<<" ";break;
				 case 9:cout<<p[9]<<" ";break;
				 default:break;
			 }
	}

    switch(d)
	{
		case 0:cout<<p[0];break;
		case 1:cout<<p[1];break;
		case 2:cout<<p[2];break;
		case 3:cout<<p[3];break;
		case 4:cout<<p[4];break;
		case 5:cout<<p[5];break;
		case 6:cout<<p[6];break;
		case 7:cout<<p[7];break;
		case 8:cout<<p[8];break;
		case 9:cout<<p[9];break;
		default:break;
	}
	return 0;
}

不知所错：

 

 

不知为何？请高人指点！多谢多谢