1123.英语数字转换器
题目地址
描述
在这个问题中,将用英语给你一个或多个整数。你的任务是将这些数字转换成整型表示。数字范围从-999,999,999到999,999,999.下面是你的程序必须考虑的详尽的英语单词表:
negative, 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, hundred, thousand, million
输入
输入包括多个样例,注意:
1.负数前面有词negative
2.当能用thousand的时候,将不用hundred。例如1500将写为"one thousand five hundred",而不是"fifteen hundred".
输入将以一个空行结束
输出
输出将是每一个单独一行,每一个后面一个换行符
样例输入
six
negative seven hundred twenty nine
one million one hundred one
eight hundred fourteen thousand twenty two
样例输出
6
-729
1000101
814022
分析
million thousand hundred 其实是三个等级,利用递归可将各级数目相加得解,而利用map可以建立英语输入和实际数字的对应关系。
反思
大神写这题只用800k,而我写了1300k,实在是化简不了了。花的时间也不短,各种调试。又犯了不少低级错误如局部变量全局变量写两次,循环括号位置错误等等。
#include<iostream>
#include<map>
#include<string>
#include<vector>
using namespace std;
map<string,int> book;
string abc[32]={"negative","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","hundred","thousand","million"};
int mindex[3]={1000000,1000,100};
int num[32]={-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,
30,40,50,60,70,80,90,100,1000,1000000};
vector<int> result;
int getn(int i,int j){ //按层次在第i和第j个数中计算和
int ans=0;
for(int p=0;p<3;++p){
for(int k=i;k<=j;++k){
if (result[k]==mindex[p]){
ans+=getn(i,k-1)*mindex[p];
i=k+1;
break;
}
}
}
for(;i<=j;++i)ans+=result[i];
return ans;
}
int main(){
for(int i=0;i<32;++i)
book.insert(make_pair(abc[i],num[i]));
while(1){
bool flag=0;
string s;
getline(cin,s); //直接输入字符串的函数getline()
int len=s.size();
if(len==0)break;
int i=0,ans=0;
for(int j=1;j<=len;++j)
if(s[j]==' '||j==len){
result.push_back(book[s.substr(i,j-i)]);
i=j+1;
j+=2;
}
if(result[0]==-1){
flag=1;
result.erase(result.begin(),result.begin()+1);
--len;
}
if(flag)cout<<'-';
cout<<getn(0,result.size()-1)<<endl;
result.clear();
}
}