PAT(A) 1082. Read Number in Chinese (25)

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

#include <cstdio>
#include <cstring>

char num[10][5]={"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
char wei[5][5]={"Shi", "Bai", "Qian", "Wan", "Yi"};

int main()
{
    char str[15];
    gets(str);                  //按字符串方式输入数字
    int len=strlen(str);        //字符串长度
    int left=0, right=len-1;    //划重点(1): left和right分别指向字符串的首位元素
    if(str[0]=='-'){            //若为负数，输出Fu，并把left右移一位
        printf("Fu");
        left++;
    }
     //划重点(2): 将right每次左移四位，直到left与right位于同一段
     //<=4为一段，省略了分段存储进二维数组的步骤
    while(left+4 <= right)     
        right -= 4;
    while(left<len){            //循环处理数字的一段（四位或小于四位）
        bool flag=false;        //表示没有积累的0
        bool isPrint=false;     //划重点(3): 表示该段没有输出过其中的位
        while(left<=right){     //从左至右处理数字中某段的每一位
            if(left>0 && str[left]=='0')    //有积累0则标记flag为true
                flag=true;
            else{                           //如果当前位不为0
                if(flag==true){             //如果存在积累的0
                    printf(" ling");
                    flag=false;
                }
                //只要不是首位（包括负号），后面的每一位前都要输出空格
                if(left>0)  printf(" ");
                printf("%s", num[str[left]-'0'] );　　//输出当前位的数字
                isPrint=true;                        //该段至少有一位被输出（即不为全0）

                if(left!=right)                      //该段除了个位，都需输出十百千
                    printf(" %s", wei[right-left-1] );   
                    //(eg!) 1234: 4-2-1=1 wei[1]=="Bai"
            }
            left++;     //划重点(4): left右移一位
        }
        if(isPrint==true && right!=len-1)            //只要不是个位，就输出万或亿
            printf(" %s", wei[(len-1-right)/4+2] );     
            //(eg) -123456789: (10-1-1)/4+2=4 wei[4]=="Yi"; 
            //                 (10-1-5)/4+2=3 wei[3]=="Wan"
        right += 4;     //划重点(5): right右移四位，输出下一段
    }
    return 0;
}