1005 Spell It Right (20)(20 分)

本文介绍了一个简单的程序设计问题:给定一个非负整数N,计算其所有数字之和,并将该和的每个数字用英文单词输出。文章提供了一段C++代码实现,能够处理不超过10^100的大整数。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

 

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

Input Specification:

Each input file contains one test case. Each case occupies one line which contains an N (<= 10^100^).

Output Specification:

For each test case, output in one line the digits of the sum in English words. There must be one space between two consecutive words, but no extra space at the end of a line.

Sample Input:

12345

Sample Output:

one five

 这题还是很简单的 但是我忘了 把0判断一下 其实这时候就体现出来了 do_while的好处了  17分

AC代码

#include<bits/stdc++.h>
using namespace std;
string num[20]={"zero","one","two","three","four","five","six","seven","eight","nine"};
stack<int>p;
int main()
{
    ios::sync_with_stdio(false);
    string s;
    while(cin>>s){
         while(!p.empty()) p.pop();
         long long int sum=0;
         for(int i=0;s[i];i++){
            sum+=s[i]-'0';
         }
        if(sum==0){
            cout<<"zero"<<endl;
            return 0;
        }

         while(sum){
              p.push(sum%10);
              sum/=10;
         }
         int flag=0;
         while(!p.empty()){
             if(flag) cout<<" ";
             cout<<num[p.top()];
             p.pop();
             flag=1;
         }
         cout<<endl;
    }
    return 0;
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值