PAT A1082 Read Number in Chinese 难_pat a1082测试点2-优快云博客

这道编程题目要求将不超过9位的整数以传统中文方式读出，包括处理负数和零的特殊情况。输入为一个整数，输出其中文读法，数字间以空格分隔。

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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:

Sample Output 2:

yi Shi Wan ling ba Bai

代码：

#include <stdio.h>
#include <string.h>

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];
    scanf("%s",str);
    int len=strlen(str);
    int left=0,right=len-1;
    if(str[0]=='-'){
        printf("Fu");
        left++;
    }
    while(left+4<=right)
        right-=4;
    while(left<len){
        bool flag=false; //有没有积累0
        bool isPrint=false;  //要不要输出位
        while(left<=right){
            if(left>0 && str[left]=='0'){
                flag=true;
            }
            else{
                if(flag){
                    printf(" ling");
                    flag=false;
                }
                isPrint=true; //该节中有一个数字不为零就要输出位（亿或万）
                if(left>0)
                    printf(" ");
                if(left<right)  //千位，百位或十位
                    printf("%s %s",num[str[left]-'0'],wei[right-left-1]);  
                else
                    printf("%s",num[str[left]-'0']);  //个位不用输出ge
            }
            left++;
        }
        if(isPrint==true && right!=len-1){
            printf(" %s",wei[(len-right)/4+2]);
        }
        right+=4;
    }
}