1082. Read Number in Chinese (25)

最新推荐文章于 2021-09-04 23:23:15 发布

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最新推荐文章于 2021-09-04 23:23:15 发布

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本文介绍了一种将负数转换为传统中文读法的方法，并提供了两种不同的C++实现方案。文章通过具体示例展示了如何处理各种特殊场景，如零的正确读法等。

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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

本题本身难度不大，但是要把所有情况考虑到比较困难，提供一些测试用例：

这是我的代码，在转换字符的时候，就控制格式，还有一别人的代码，是在转换完毕后，再控制输出格式。

#include<iostream>
#include<string>
#include<vector>
using namespace std;
string num[10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
string jinwei[6]={"","Shi","Bai","Qian","Wan","Yi"};
vector<string> out;
void changetos(int n){
    int k=0;
    bool first0=true,get0=false;//判断是否为首位0，连续0
    while(n!=0){
        for(int i=0;i<=3;i++){
            int t=n%10;
            if(t==0)
            {
                if(!first0&&!get0){
                    out.push_back(num[t]);
                }
                get0=true;
            }
            else{
                if(i!=0){
                        out.push_back(num[t]+" "+jinwei[i]);
                }
                else out.push_back(num[t]);
                get0=false;
                first0=false;
            }
            n/=10;
            if(n==0) break;
            k++; 
            if(k==4||k==8) {//判断是否是万位或亿位
                if(k==8&&out[out.size()-1]=="Wan")
                    out.erase(out.end()-1); 
                out.push_back(jinwei[4+k/8]);first0=true;//重置首位0
            }
        }
    }
}
int main(){
    freopen("in.txt","r",stdin);
    int n;
    cin>>n;
    if(n<0) {
        cout<<"Fu ";
        n=-n;
    }
    if(n){
        changetos(n);
        for(int i=out.size()-1;i>0;i--){
            cout<<out[i]<<" ";
        }
        cout<<out[0];
    }
    else cout<<num[0];

    return 0;
}

感谢小5555的分享。

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

int main()
{
    string a[] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
    string b[] = {"", "Shi", "Bai", "Qian", "Wan", "Shi", "Bai", "Qian", "Yi"};
    vector<string> res;
    vector<int> digit;
    int num, e;
    cin >> num;
    if (num == 0){
        cout << "ling";
        return 0;
    } 
    else if (num < 0){
        cout << "Fu ";
        num = -num;
    }
    while (num != 0){
        digit.push_back(num%10);
        num /= 10;
    }
    for (e=0; e<digit.size() && digit[e]==0; ++ e) {}
    if (e == 8){
        cout << a[digit[e]] << " Yi";
        return 0;
    }
    for (int i = e; i < digit.size(); ++ i){
        if (i!=0 && (digit[i]!=0 || i==4 || i==8)){
            res.push_back( b[i] );
        }
        res.push_back(a[ digit[i] ]);
    }
    for (int i = res.size()-1; i >= 0; -- i){
        if (i != res.size()-1){
            cout << " ";
        }
        int cnt = 0;
        while (i>=0 && res[i]=="ling"){
            -- i;
            ++ cnt;
        }
        if (cnt!=0 && res[i]!="Wan"){
            cout << "ling ";
        }
        cout << res[i];
    }

    return 0;
}