PTA basic question 7-23 currency conversion (20 points) (true)

7-23 Currency conversion (20 branch )

Enter an integer （ The number of digits does not exceed 9 position ） Represents a RMB value （ The unit is yuan ）, Please convert to the capital Chinese format required by the financial department . Such as 23108 element , Transformed into “ Twenty three thousand one hundred and eight ” element . To simplify the output , In lowercase a-j The order represents capital numbers 0-9, use S、B、Q、W、Y Each represents ten 、 hundred 、 Thousand 、 ten thousand 、 Billion . therefore 23108 The element should be converted to output “cWdQbBai” element .

Input format ：

The input gives no more than 9 A nonnegative integer of bits .

Output format ：

Output the converted result in one line . Be careful “ zero ” The usage of must conform to Chinese habits .

sample input 1：

813227345

No blank lines at the end

sample output 1：

iYbQdBcScWhQdBeSf

No blank lines at the end

sample input 2：

6900

sample output 2：

gQjB

analysis ： This question is true. It's disgusting , Many special situations and details , I haven't thought of a good method for the time being , Poke an eye , Update later if you have ideas . The current idea is  ： Start with the most general situation , for example 123456789, Then consider the middle and end bands 0 The situation of ; The code is as follows ：

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int a[10],cnt=0;
    do {
        a[cnt++]=n%10;
        n/=10;
    } while (n);
    char b[15]="abcdefghij";
    int t=0;
    for (int i=cnt-1;i>=t;i--) {
    	if (i==4&&a[i]==0) {
    		printf("W");
    		continue;
		} else printf("%c",b[a[i]]); // Wan weishang is 0 No output 0, Ten thousand still need to be output , example 6902312;
        // 6900;  Handle the ending band 0（ Do not output all ending bands 0 Part of , When 10000 bits are included, 10000 still need to be output ）
		if (a[0]==0) {
        	int j=1;
        	while (a[j]==0) {
        		j++;
			}
			if (j>=1) t=j;
		}
    	// 100000001; 69002 Deal with the median 0, Push it down  （ Multiple zeros are output only once ）
    	if (a[i]==0) {
    		int j=i;
    		while (a[j]==0) {
    			j--;
			}
			i=j+1;
		}
        if (i==4) printf("W");
		if (a[i]) {
			if (i==8) printf("Y");
        	if (i==7||i==3) printf("Q");
       		if (i==6||i==2) printf("B");
        	if (i==5||i==1) printf("S");
		}
    }
}

Debugging slowly can only be said to be , There are also various ways to solve this problem , Poke an eye , Let's see later .