字符串-POJ/UVA 486-English-Number Translator(英语数字转换器)

Dear 丶 2022-05-14 22:29 166阅读 0赞

*  # 字符串-POJ/UVA 486-English-Number Translator(英语数字转换器) #

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*  ## 题目链接：[5:英语数字转换器][5] ##
 *  ## 思路： ##

有中文题目不讲大意

这道题会卡人的就是奇葩的英语表示数字大小，外国million下来直接就thousand，所以会出现million和thousand之间隔着个--nine hundred ninty nine ，其实无非就三种情况需要考虑这个问题，遇到million，thousand，hundred，需要回头计算乘上这些表示它们个数的数字，而且也容易发现，thousand 前面的倍数肯定不会比它自己大（因为再大就百万了），hundred也类似

***所以取余是一个好东西，只要遇到 million，thousand，hundred 就对前面的计算结果取余，然后前面的结果减掉这部分余数，再加上 million，thousand，hundred乘上这个余数的值***

英文数字用多个map储存

我是用多个map，为了表示出优先级，位数越大优先级越高，便于转换

*  ## 代码： ##

#include<iostream>	
    #include<sstream>	
    #include<string>
    #include<map>
    #include<queue>	
    using namespace std;
    
    string P1[] = { "zero","one","two","three","four","five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen","fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
    string P2[] = { "ten","twenty", "thirty", "forty", "fifty","sixty", "seventy", "eighty", "ninety" };
    string P3[] = { "hundred", "thousand", "million" };
    string Wait_Op_num;
    int Mod[] = { 100,100, 10,1000,1000000};
    map<string,int> Num_Set[5];
    
    void Pre_Op()   //构建map
    {
    	for (int i = 0; i < 20; i++)
    		Num_Set[0].insert(pair<string, int>(P1[i], i));
    	for (int i = 1; i < 9; i++)
    		Num_Set[1].insert(pair<string, int>(P2[i], 10 * (i + 1)));
    	Num_Set[2].insert(pair<string, int>(P3[0], 100));
    	Num_Set[3].insert(pair<string, int>(P3[1], 1000));
    	Num_Set[4].insert(pair<string, int>(P3[2], 1000000));
    
    }
    
    int main()
    {
    	Pre_Op();
    	while (getline(cin, Wait_Op_num))
    	{
    		if (Wait_Op_num.empty())
    			break;
    		int Res = 0;
    		int Privelige = -1;
    		int Temp_Pre;
    		int Negative_flag = 0;
    		stringstream ss(Wait_Op_num);   //字符串流
    		queue<string> Num;
    		while (!ss.eof())
    		{
    			string temp_str;
    			ss >> temp_str;
    			//cout << temp_str << endl;
    			Num.push(temp_str);   //单词入队列
    		}
    		ss.clear();
    
    		while (!Num.empty())
    		{
    			int Check_Index = 0;
    			int Number;
    			if (Num.front() == "negative")   //负数标记
    			{
    				Num.pop();
    				Negative_flag = 1;
    			}
    			string Front = Num.front();
    			Num.pop();
    
    			for (Check_Index = 0;; Check_Index++)   //对单词进行数值转换
    			{
    				auto it = Num_Set[Check_Index].find(Front);
    				if (it!= Num_Set[Check_Index].end())
    				{
    					Number = it->second;
    					break;
    				}
    			}
    			if (Privelige == -1)   //优先级为-1，接触第一个数字
    			{
    				Res += Number;
    				Privelige = Check_Index;
    			}
    			else if (Privelige > Check_Index)   //当前单词小于最大优先级
    			{
    				if (Check_Index >= 2)   //遇到 thousand hundred million
    				{
    					//one million twenty one thousand one hundred one
    					int Tail_num = Res % Mod[Check_Index];   //取余
    					Res -= Tail_num;   //减去余数
    					Res += Tail_num * Number;   /加上
    				}
    				else
    					Res += Number;   //如果是十位数，个位数 直接加上去
    			}
    			else if (Privelige < Check_Index)  //当前数优先级高
    			{
    				Res *= Number;
    				Privelige = Check_Index;
    			}
    			//cout << Res << endl;
    		}
    		if (Negative_flag)
    			Res *= -1;
    		cout << Res << endl;
    	}
    	return 0;
    }
    
    //negative nine hundred ninety nine million nine hundred ninety nine thousand nine hundred ninety nine

[5]: http://dsalgo.openjudge.cn/string/5/