字符串

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

• 题目链接：5:英语数字转换器

• 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 oneint Tail_num = Res % Mod[Check_Index];   //取余Res -= Tail_num;   //减去余数Res += Tail_num * Number;   /加上}elseRes += 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