POJ 2121 | 易学教程

Inglish-Number Translator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5162		Accepted: 2027

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

C++string类写

#include <iostream>
#include <string>
#include <cstdio>

using namespace std;



string a[100]={"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 main(){
string s;
int ans=0,sum=0;
while(cin>>s){
    if(s==a[0])
        cout<<"-";
    else if(s==a[1])
        sum+=0;
    else if(s==a[2])
        sum+=1;
     else if(s==a[3])
        sum+=2;
        else if(s==a[4])
        sum+=3;
        else if(s==a[5])
        sum+=4;
        else if(s==a[6])
        sum+=5;
        else if(s==a[7])
        sum+=6;
        else if(s==a[8])
        sum+=7;
        else if(s==a[9])
        sum+=8;
        else if(s==a[10])
        sum+=9;
        else if(s==a[11])
        sum+=10;
        else if(s==a[12])
        sum+=11;
        else if(s==a[13])
        sum+=12;
        else if(s==a[14])
        sum+=13;
        else if(s==a[15])
        sum+=14;
        else if(s==a[16])
        sum+=15;
        else if(s==a[17])
        sum+=16;
        else if(s==a[18])
        sum+=17;
        else if(s==a[19])
        sum+=18;
        else if(s==a[20])
        sum+=19;
        else if(s==a[21])
        sum+=20;
        else if(s==a[22])
        sum+=30;
        else if(s==a[23])
        sum+=40;
        else if(s==a[24])
        sum+=50;
        else if(s==a[25])
        sum+=60;
        else if(s==a[26])
        sum+=70;
        else if(s==a[27])
        sum+=80;
        else if(s==a[28])
        sum+=90;

    else if(s==a[29])
        sum*=100;
    else if(s==a[30])
    {
        ans+=sum*1000;
        sum=0;
    }
    else if(s==a[31])
    {
        ans+=sum*1000000;
        sum=0;
    }
    char ch = getchar();
    if(ch=='\n'){
        cout<<sum+ans<<endl;
        sum=0;
        ans=0;
    }
}
return 0;
}

C的strcmp写

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 100
char s[maxn];
int sum=0,ans=0;
int main()
{
 while(scanf("%s",s)!=EOF)
 {
  if(!strcmp(s,"negative"))printf("-");
  else if(!strcmp(s,"zero"))sum+=0;
  else if(!strcmp(s,"one"))sum+=1;
  else if(!strcmp(s,"two"))sum+=2;
  else if(!strcmp(s,"three"))sum+=3;
  else if(!strcmp(s,"four"))sum+=4;
  else if(!strcmp(s,"five"))sum+=5;
  else if(!strcmp(s,"six"))sum+=6;
  else if(!strcmp(s,"seven"))sum+=7;
  else if(!strcmp(s,"eight"))sum+=8;
  else if(!strcmp(s,"nine"))sum+=9;
  else if(!strcmp(s,"ten"))sum+=10;
  else if(!strcmp(s,"eleven"))sum+=11;
  else if(!strcmp(s,"twelve"))sum+=12;
  else if(!strcmp(s,"thirteen"))sum+=13;
  else if(!strcmp(s,"fourteen"))sum+=14;
  else if(!strcmp(s,"fifteen"))sum+=15;
  else if(!strcmp(s,"sixteen"))sum+=16;
  else if(!strcmp(s,"seventeen"))sum+=17;
  else if(!strcmp(s,"eighteen"))sum+=18;
  else if(!strcmp(s,"nineteen"))sum+=19;
  else if(!strcmp(s,"twenty"))sum+=20;
  else if(!strcmp(s,"thirty"))sum+=30;
  else if(!strcmp(s,"forty"))sum+=40;
  else if(!strcmp(s,"fifty"))sum+=50;
  else if(!strcmp(s,"sixty"))sum+=60;
  else if(!strcmp(s,"seventy"))sum+=70;
  else if(!strcmp(s,"eighty"))sum+=80;
  else if(!strcmp(s,"ninety"))sum+=90;
  else if(!strcmp(s,"hundred"))
  {
    sum*=100;
  }
  else if(!strcmp(s,"thousand"))
  {
   ans+=sum*1000;<pre><span style="color:#008000;">//</span><span style="color:#008000;">当为1000之后，原数应加上num*1000，而非直接乘，1000000也是一样</span>

sum=0; } else if(!strcmp(s,"million")) { ans+=sum*1000000; sum=0; }

//当系统读到回车时，输出 

 char ch=getchar(); 
if(ch=='\n') 
{
 printf("%d\n",ans+sum);
 sum=0; ans=0;
 } 
}
 return 0;
}

来源：https://www.cnblogs.com/mingrigongchang/p/6246284.html

标签

strcmp

negative