PAT1082. Read Number in Chinese

Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way.  Output "Fu" first if it is negative.  For example, -123456789 is read as "Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu".  Note: zero ("ling") must be handled correctly according to the Chinese tradition.  For example, 100800 is "yi Shi Wan ling ba Bai".

Input Specification:

Each input file contains one test case, which gives an integer with no more than 9 digits.

Output Specification:

For each test case, print in a line the Chinese way of reading the number.  The characters are separated by a space and there must be no extra space at the end of the line.

Sample Input 1:

-123456789

Sample Output 1:

Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu

Sample Input 2:

100800

Sample Output 2:

yi Shi Wan ling ba Bai
思路：比较麻烦的题 需要考虑多种情况。  四个四个进行处理 并且需要考虑到0的情况

 1 #include<cstdio>
 2 #include<cstring>
 3 char data[10][6]=
 4 {
 5     "ling",
 6     "yi",
 7     "er",
 8     "san",
 9     "si",
10     "wu",
11     "liu",
12     "qi",
13     "ba",
14     "jiu"
15 };
16 char sig[6][9]=
17 {
18     "Shi",
19     "Bai",
20     "Qian",
21     "Wan",
22     "Yi"
23 };
24 char sen[11];
25 int main(int argc, char *argv[])
26 {
27     scanf("%s",sen);
28     //注意处理0的问题
29     int len=strlen(sen);
30     int left=0,right=len-1;
31     if(sen[0]==‘-‘)
32     {
33         left++;
34         printf("Fu");
35     }
36     while(left+4<=right)
37     {
38         right-=4;
39     }
40     while(left<len)
41     {
42         bool accumalte=false;
43         bool Isprint=false;
44         while(left<=right)
45         {
46             if(left>0&&sen[left]==‘0‘)
47             {
48                 accumalte=true;
49             }
50             else
51             {
52                  if(accumalte)
53                  {
54                     accumalte=false;
55                     printf(" ling");
56                 }
57                 if(left>0)
58                    printf(" ");
59                   printf("%s",data[sen[left]-‘0‘]);
60                   Isprint=true;
61                   if(left!=right)
62                         printf(" %s",sig[right-left-1]);
63             }
64             left++;
65         }
66         if(Isprint&&right!=len-1)
67         {
68             printf(" %s",sig[(len-right-1)/4+2]);
69         }
70         right+=4;
71     }
72     putchar(‘\n‘);
73     return 0;
74 }