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  • 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 }
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  • 原文地址:https://www.cnblogs.com/mld-code-life/p/11218832.html
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