poj 2121 Inglish-Number Translator 将英文翻译成罗马数字

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该博客主要介绍了如何解决POJ 2121问题，即从英文形式的数字转换为罗马数字。文章中提到，输入的负数会带有“negative”前缀，并且在可以使用“thousand”的情况下不会使用“hundred”。文章未给出具体的解决方案，但预期会包含转换的策略和步骤。

Inglish-Number Translator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2822		Accepted: 1107

Description

In this problem, you will be given one or more integers in English. Your task is to translate these numbers into their integer representation. The numbers can range from negative 999,999,999 to positive 999,999,999. The following is an exhaustive list of English words that your program must account for:
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

Input

The input consists of several instances. Notes on input:

Negative numbers will be preceded by the word negative.
The word "hundred" is not used when "thousand" could be. For example, 1500 is written "one thousand five hundred", not "fifteen hundred".

The input is terminated by an empty line.

Output

The answers are expected to be on separate lines with a newline after each.

Sample Input

six
negative seven hundred twenty nine
one million one hundred one
eight hundred fourteen thousand twenty two

Sample Output

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<sstream>
using namespace std;
string sig="negative";
string tmp[31]={"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 position(string str)
{
    for(int i=0;i<31;i++) if(tmp[i]==str) return i;
}
int val(int n)
{
    if(n>=20&&n<=27) return (n-18)*10;
    else if(n==28) return 100;
    else if(n==29) return 1000;
    else if(n==30) return 1000000;
    return n;
}
int main()
{
    string s;
    while(getline(cin,s),s!="")
    {
        istringstream sin(s);
        string str[100];
        int len=0;
        while(sin>>str[len]) len++;
        int cnt=0,tmp=0;
        //不会出现one thousand millon 这种情况
        for(int i=0;i<=len;i++)
        {
            if(i==len) {cnt+=tmp;break;}
            if(str[i]==sig) {printf("-");continue;}
            int tag=val(position(str[i]));
            if(tag==100) tmp*=tag;
            else if(tag>100) tmp*=tag,cnt+=tmp,tmp=0;
            else tmp+=tag;
        }
        printf("%d/n",cnt);
    }
    return 0;
}