【PAT (Advanced Level)】1005. Spell It Right (20)

解读编程技巧与技术领域

最新推荐文章于 2019-02-12 18:27:53 发布

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1005. Spell It Right (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<= 10¹⁰⁰).

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:

Sample Output:

one five

题目意思明确，输入肯定不能用数值、只能用数组或者字符串，然后把输入按位相加，把结果按位输出就好了，很简单；

代码：

/************************************
This is a test program for anything
Enjoy IT!
************************************/

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <list>
#include <set>
#include <fstream>
#include <string>
#include <cmath>
using namespace std;

int main()
{
	/*ifstream cin;
	cin.open("in.txt");
	ofstream cout;
	cout.open("out.txt");*/
	//////////////////////////////////////////////////////////////////////////
	// TO DO Whatever You WANT!
	string s1;
	while (cin >> s1 )
	{
		int sum = 0;
		list<string> out;
		for (int i = 0; i < s1.length(); i++)
		{
			sum += s1[i] - '0';
		}
		do 
		{
				switch (sum % 10)
				{
				case 0:
					out.push_front("zero");
					break;
				case 1:
					out.push_front("one");
					break;
				case 2:
					out.push_front("two");
					break;
				case 3:
					out.push_front("three");
					break;
				case 4:
					out.push_front("four");
					break;
				case 5:
					out.push_front("five");
					break;
				case 6:
					out.push_front("six");
					break;
				case 7:
					out.push_front("seven");
					break;
				case 8:
					out.push_front("eight");
					break;
				case 9:
					out.push_front("nine");
					break;
				}
				sum /= 10;
		} while (sum != 0);
		int len = out.size();
		for (int i = 0; i < len; i++)
		{
			if (i == 0)
			{
				cout << out.front();
			}
			else
			{
				cout << " " << out.front();
			}
				out.pop_front();
		}
		cout << endl;
	}
	//////////////////////////////////////////////////////////////////////////
	// system("pause");
	return 0;
}