1082 Read Number in Chinese （25 分）

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

参考代码：

#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"};
const int MANX = 15;
char s[MANX];
int main(int argc, char const *argv[])
{
  fgets(s, 16, stdin);
  int len = strlen(s) - 1;//fgets会多读入一个换行符
  int left = 0;
  int right = len - 1;

  if(s[0] == '-'){
    printf("Fu");
    left++;
  }

  while(left + 4 <= right){//让left和right在同一字节，四位为一个字节
    right -= 4;
  }

  int count = 0;
  while(left < len){
    bool flag = false;//有无累计字节
    bool isPrintf = false;//该字节有无输出
    while(left <= right){//从左向右输出字节每一位
      if(s[left] == '0' && left > 0){//left > 0 是为了当只有一个零的时候，输出零
        flag = true;
        count++;//统计有无一个字节四个都为零的情况
      }else{
        if(flag == true ){//如果存在累计零
          printf(" ling");
          flag = false;
        }
        if(left > 0) printf(" ");//不是第一位，则输出空格
        printf("%s", num[s[left] - '0']);//输出当前位
        isPrintf = true;//该节至少有一个被输出
        if(right != left){//该节除了个位外，都要输出十百千
          printf(" %s", wei[right - left - 1]);
        }
      }
      left++;//右移一位
    }
    if(isPrintf == true && right != len - 1){//该字节有输出，且只要不是个位，就输出万或者亿
      printf(" %s", wei[(len - right - 1) / 4 + 2]);
    }
    if(count == 4 && s[right + 1] != '0' && right != len - 1){//当上一个字节都为零，该字节第一个不为零时，输出零
      printf(" ling");
    }
    right += 4;
    count = 0;
  }
  printf("\n");
  return 0;
}