PTA第二题

 1 #include<string.h>
 2 #include<stdio.h>
 3 #include<malloc.h>
 4 
 5 char n[120];
 6 char a[10][5]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
 7 char *p[10];
 8 int main()
 9 {
10     scanf("%s",n);               //大数转换成字符串处理
11     int len=strlen(n), sum=0;
12 
13     for(int i=0;i<len;i++)
14         sum+=((int)n[i]-48);
15 
16     int j,k=0;
17     if(sum==0){
18         printf("ling");
19         return 0;
20     }
21 
22     while(sum!=0){
23         j=sum%10;
24         for(int i=0;i<10;i++){
25             if(j==i){
26                  p[k++]=a[i];
27                  break;
28             }
29         }
30         sum=sum/10;
31     }
32 
33     for(int i=k-1;i>0;i--)
34         printf("%s ",p[i]);
35 
36     printf("%s",p[0]);
37 
38     return 0;
39 }

 

posted @ 2019-07-20 18:59  三行代码划江湖  阅读(189)  评论(0编辑  收藏  举报