1082 Read Number in Chinese （字符串

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

分析

本题的难点在于如何处理零，自己的代码在处理零时总存在测试点通不过，参考博客^[1],现在对博客的解法做一点个人的理解分析。解法使用zero标记是否在下一次碰到非零元素时是否输出合理的零（i.e.连续0或者单独一个0，按照中文读法输出一个ling），当zero=false时表示还未遇见0或者上次的0事件（i.e.连续0或者单独0）已经处理完毕，当zero=true时，下次遇到非零元素时需要处理0，即先输出ling再输出非零元素的中文读法，或者一直到最后一位（个位）依然未出现0元素，则不输出ling，zero=true在本次标记中无效。

#include <iostream>
#include <algorithm>
#include<vector>
#include<cmath>
using namespace std;
string kv[]= {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
string bit[]= {"","Shi","Bai","Qian"};
string bit2[]= {"Yi","Wan",""};
int main() {
    int n;
    cin>>n;
    if(n==0) {
        cout<<"ling";
        return 0;
    }
    if(n<0) {
        cout<<"Fu ";
        n=-n;
    }
    int part[3];
    part[0]=n/100000000,part[1]=(n-n/100000000*100000000)/10000,part[2]=n%10000;
    bool zero=false;
    int printCnt=0;
    for(int i=0; i<3; i++) {
        int tmp=part[i];
        for(int j=3; j>=0; j--) {
            int curPos=8-i*4+j;
            if(curPos>=9)  continue;
            int cur=(tmp/(int)pow(10,j))%10;
            if(cur!=0) {
                if(zero) {
                    cout<<" ling";
                    zero=false;
                }
                if(j==0) {
                    printCnt++==0?cout<<kv[cur]:cout<<" "<<kv[cur];
                } else {
                    printCnt++==0?cout<<kv[cur]<<" "<<bit[j]:cout<<" "<<kv[cur]<<" "<<bit[j];
                }
            } else {
                if(!zero&&j!=0&&n/pow(10,curPos)>=10) zero=true;
            }
        }
        if(i!=2&&part[i]>0) cout<<" "<<bit2[i];
    }
    return 0;
}

1. https://www.liuchuo.net/archives/2204 ↩