[刷题] 1002 写出这个数 (20分)

要求

• 读入一个正整数 n，计算其各位数字之和，用汉语拼音写出和的每一位数字

输入格式

• 每个测试输入包含 1 个测试用例，即给出自然数 n 的值。这里保证 n 小于 1

输出格式

• 在一行内输出 n 的各位数字之和的每一位，拼音数字间有 1 空格，但一行中最后一个拼音数字后没有空格

输入样例

• 1234567890987654321123456789

输出样例

• yi san wu

实现

• java

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner in = new Scanner(System.in);
        String[] strings = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
        
        String s = in.nextLine();
        char[] chars = s.toCharArray();
        int sum = 0;
        for(char c:chars){
            int x = c - '0';
            sum += x;
        }
        String sumS = String.valueOf(sum);
        char[] charSum = sumS.toCharArray();
        StringBuilder StringBuilder = new StringBuilder();
        for(char c:charSum){
            int x = c - '0';
            StringBuilder.append(strings[x]+" ");
        }
        StringBuilder.deleteCharAt(StringBuilder.length()-1);
        System.out.println(StringBuilder.toString());
    }
}

• c++

#include<iostream>
#include<string>
using namespace std;
int main(){
    string str;
    getline(cin,str);
    int sum=0;
    for(int i = 0; i < str.size(); i ++)
        sum += str[i] - '0';
    string a[10] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
    string re = to_string(sum);
    for( int i = 0 ; i < re.size() - 1 ; i ++ )
        cout << a[re[i] - '0'] << " ";
    cout << a[[re.size()-1]-'0'];
}