EASY_ZJU_PAT_ADVANCED LEVEL 1005

解读编程挑战：数字转英文单词之和

最新推荐文章于 2014-03-06 19:24:16 发布

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本文深入解析编程挑战题目，指导如何将数字转换为英文单词表示之和，涉及字符串操作、数字分解与转换等核心算法，提供详细步骤与代码实现，帮助开发者解决实际编程问题。

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

/***************************************************88
	
	@ AUTHOR: GAOMINQUAN
	@ MAIL	: ENSOLEILLY@GMAIL.COM
	@ DATA	: 2014 - 2 - 22

**************************************************/

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


string words[11] = {"zero","one","two","three","four","five","six","seven","eight","nine","ten"};
int digitSum(string num){
	int sum = 0;
	for(int stringIndex = 0; stringIndex <num.length(); stringIndex++){
		sum += num[stringIndex] - '0';
	}
	return sum;
}
void split(int num){
	
	vector<int> splitNum;
	if( num == 0){
		splitNum.push_back(0);
	}else{
		while(num>0){
			int tail = num%10;
			splitNum.push_back(tail);
			num /= 10;
		}
	}
	
	for(int splitIndex = splitNum.size() - 1; splitIndex > 0; splitIndex--){
		cout<<words[splitNum[splitIndex]]<<" ";
	}cout<<words[splitNum[0]];
}

int main(){
	
	string num = "1";
	cin>>num;
	int sum = digitSum(num);
	
	split(sum);

	return 0;
}
	
NOTE：

D:\MSDev98\Bin\CODE\PAT_AL_1005.CPP(34) : error C2679: binary '<<' : no operator defined which takes a right-hand operand of type 'class std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >' (or there is no acceptable co


1. 在使用string的时候，要记得把string先导入才行；
2. 数字非常长的时候，不能用int
3. 位数分解的时候，要考虑零的情况。