A-1082 Read Number in Chinese (25 分)

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本文介绍了一个C++程序，用于将不超过9位的整数转换为传统中文读法。该程序能够正确处理负数和零的情况，例如-123456789读作“负一亿二千三百四十五万六千七百八十九”。文章通过实例演示了如何实现这一功能。

A-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

#include <string>
#include <iostream>
using namespace std;

int main(){
    int m,n,i,j,k,x;
    int w1[9]={},w[9]={};
    string w2[10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
    string s[9]={{},"Shi","Bai","Qian","Wan","Shi","Bai","Qian","Yi"};
    cin >> n;
    if(n<0){
        cout << "Fu ";
        n=-n;
    }
    m=n;
    if(m<10){
        cout << w2[m];
        return 0;
    }
    for(i=0;m>0;i++){
        w[i]=m%10;
        m/=10;
    }
    for(j=i-1;j>=0;j--){
        if(w[j]!=0&&j==i-1){//对第一个数字、最后一个数字和中间数字的空格的处理不同
            cout << w2[w[j]] << ' ' << s[j];
        }else if(w[j]!=0&&j>0&&j<i-1){
            cout << ' ' <<w2[w[j]] << ' ' << s[j];
        }else if(w[j]!=0&&j==0){
            cout << ' '<<w2[w[j]];
        }
        if(w[j]==0){
            x=0;
            while(w[j]==0){//删去中间连续的0，并记录个数
                j--;
                x++;
            }
            if(j+x>3&&j+x<9&&j<=3)cout << " Wan";//如果删除之前的数在‘万’以上，并且在删除之后在‘万’以下就输出一个‘万’
            if(w[j+1]==0&&j>=0&&w[j]!=0)cout << " ling";//两个数中间有0，就输出一个‘ling’
            j++;
        }
    }
    
    return 0;
}