OPENJUDGE 3713 外星人翻译用数字模块

最新推荐文章于 2021-11-13 13:29:29 发布

原创最新推荐文章于 2021-11-13 13:29:29 发布 · 367 阅读

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#include #stdstring #iostream #dictionary #structure #iterator

本文介绍了一种数值到词语的高效转换算法，通过建立字典、使用堆栈操作和逐级匹配，实现数字到中文词语的精确转换。算法特别考虑了千位、万位等进位规则，以及负数的特殊处理。

Algorithm Abstract:

1)Establishing a Look_up Table As A Dictionary with STL map Structure;

2)Get Input And Push Items Into A Stack;

3)Pop Items In The Stack And Compare with items in the dictionary

We locate each number by a Section-Offset Algorithms, when "thousand", “Million” Are received from the stack, the exponent factor were set 3/6, when "hundred" is received, exponent += 2(which is the offset in each 3-digit section)

Caution:

1) Remember to use clear() every time u need to restore the stringstream object

2) eof() returns TRUE when the tail of stream object(sstream,fstream,iostream) is

# include <iostream>
# include <sstream>
# include <map>
# include <stack>
# include <vector>
# include <string>
# include <cmath>

using namespace std;

string loader1[29] = 

{"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","thirteen","fourteen"

,"fifteen","sixteen","seventeen","eighteen","nineteen","twenty","thirty","forty","fifty","sixty","seventy","eighty",

"ninety","*"};
int loader2[29] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,-1};

map<string,int> Dictionary;

void initializeDic()
{
	for ( int i = 0; i < 29; i++ )
	{
		Dictionary.insert(make_pair(loader1[i],loader2[i]));
	}
}

int power10(int f)
{
	int result = 0;
	double exp = pow(10,(double)f);
	result = (int)exp;
	return exp;
}

int main()
{
	
	initializeDic();
	string inputString;
	stringstream s;
	
	string word;
	bool flag = false;
	int factor = 0;
	int sum = 0;
	stack<string> numStack;
	map<string,int>::iterator itr;

	while( getline(cin,inputString) )
	{
		s.clear();
		s.str(inputString);
		s.seekg(ios::beg);	//Don't Forget This!
		factor = 0;
		sum = 0;
		flag = false;

		while ( !s.eof() )     //eof was defined as the tail of the string
		{
			s >> word;
		
			numStack.push(word);	
			word.clear();
		}
		while ( !numStack.empty() )
		{
			word = numStack.top();
			if ( Dictionary.find(word) != Dictionary.end() )
			{
				itr = Dictionary.find(word);
				sum += itr->second * power10(factor);
			}
			else if ( word == "hundred" )
			{
				factor += 2;
			}
			else if ( word == "thousand" )
			{
				factor = 3;
			}
			else if ( word == "million" )
			{
				factor = 6;
			}
			else if ( word == "negative" )
			{
				flag = true;
			}
			numStack.pop();
		}
		if ( flag )
		{
			cout << -sum << endl;
		}
		else
		{
			cout << sum << endl;
		}
	}
	return 0;
}