7-23 currency conversion (using array conversion)

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

sample output 1：

iYbQdBcScWhQdBeSf

sample input 2：

6900

sample output 2：

gQjB

Solution 1 ：

#include <stdio.h>
int main()
{
	int i=0,j,n;
	scanf("%d",&n);
	char a[9];
	char b[9]={'\n','S','B','Q','W','S','B','Q','Y'};  // The names of bits are stored in the array b
	if(n==0){
		printf("a\n");
	} else{
		while(n!=0){   // Numbers are stored in arrays a
		a[i]=n%10;
		n=n/10;
		i++;
		}
		i=i-1;
		for(;i>=0;i--){            // Array a Reverse output 
			if(a[i]==0){           
				if(i==4){               // Wan Weiwei 0 The situation of 
					if(a[8]!=0 && a[7]+a[6]+a[5]+a[4]==0  ){   // There are 100 million , And all ten thousand bits are zero 
						printf("a");                           // Output a zero （ Such as 100004800）
					}else{
						printf("W");                           // No, if there are ten thousand bits on the side, one ten thousand will be output .
					}
				}else{
					if(a[i-1]==0){                              // repeat 0 No output 
					continue;
					}else{                                 //  General 0 Output zero , No bit name 
						printf("a");                
					}
				}	
			}else{
				printf("%c%c",a[i]+'a',b[i]);
			}	
		}
	}
	
	return 0;
}

summary ：

There is an array of numbers , There is an array of digit names . Reverse output two arrays from large to small .

The other is the determination of reading rules .

Solution 2 ：

#include <stdio.h>
#include <string.h>
int main()
{
	int i,n;
	char a[15];
	scanf("%s",&a);
	char b[9]={'\n','S','B','Q','W','S','B','Q','Y'};
    if(a[0]=='0') {printf("a\n"); return 0;}
    n=strlen(a);
    for(i=0;i<n;i++){  
        if(a[i]=='0'){           
            if(i==n-4) printf("W");
            if(i<n-1&&a[i+1]=='0'||i==n-1)continue; // The pronunciation of Chinese zero 
            else printf("a");
        }else  printf("%c%c",a[i]+'a'-'0',b[n-i-1]);
	}
	return 0;
}

Input the number as a string （ Forward input ）