A + B Again

A + B Again

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 201, Accepted users: 182

Problem 10281 : No special judgement

Problem description

Reads in positive integers A and B,calculates A + B.Pay attention:A and B is composed by simple English word. [0-9]:"zero","one","two","three","four","five","six","seven","eight","nine";

Input

Standard input will contain multiple test cases.Each testcase occupies a line.the form is "A + B =",the neighboring two characters string has a blank space.When A and B is 0 at the same time ,inputs finished.You shouldn't output the result 0.

Output

For each testcase outputs 1 line,The result of A + B,by simple English word too.Each word is separated with a single space.There's no space after the last word.

Sample Input

one + two = three four + five six = zero + zero =

Sample Output

three nine zero

Problem Source

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本人自己的解法,相当麻烦:

 

  
#include<iostream> 
#include<string> 
#include<map> 
using namespace std;
int pow(int x, int y)
{
     int sum = x;
     for(int i=0; i<y; i++)
     {
         sum*=10;
     }
     return sum;
}
int main(int argc, char* argv[])
{    
    int i = 0;
    int len;
    int r=0;
    int l=0;
    char *ptr;
    int t[1000];
    string*s = new string[1000];
    map<string,int> m;
    map<int,string> m1;
    m["zero"] = 0;
    m["one"] = 1;
    m["two"] = 2;
    m["three"] = 3;
    m["four"] = 4;
    m["five"] = 5;
    m["six"] = 6;
    m["seven"] = 7;
    m["eight"] = 8;
    m["nine"] = 9;
    m1[0] = "zero" ;
    m1[1] = "one";
    m1[2] = "two";
    m1[3] = "three";
    m1[4] = "four";
    m1[5] = "five";
    m1[6] = "six";
    m1[7] = "seven";
    m1[8] = "eight";
    m1[9] = "nine";
    
    char in[100000];
    while(gets(in))
    {
       len = strlen(in)+1;
       ptr = strtok(in, " ");
       while(ptr)
       {
         if(ptr[0]!='+'&&ptr[0]!='=')
         {
            t[i] = m[ptr];
            i++;
         }
         else
         {
            
             for(int k=i-1,f=0; k>=0; k--,f++)
             {
                 r+=pow(t[f],k);
            
             }
            
             i=0;
         }
         ptr = strtok(NULL," ");
       }
    
       if(r == 0)
           return 0;
       int temp = r;
       int count = 0;
       while(r)
       {
          s[count++]=m1[r%10];
          
          r = r/10;
          l++;
        
       } 
       for(int p=count-1; p>=0; p--)
       {
           cout<<s[p];
           if(p>0)
              cout<<" ";
       }
       cout<<endl;
        
     
       l=0;
       r=0;
      
    }
    
    return 0;
}
/*在其中学会了如何使用一个函数:
原型：extern char *strtok(char *s, char *delim);
        
  用法：#include <string.h>
  
  功能：分解字符串为一组标记串。s为要分解的字符串，delim为分隔符字符串。
  
  说明：首次调用时，s必须指向要分解的字符串，随后调用要把s设成NULL。
        strtok在s中查找包含在delim中的字符并用NULL('/0')来替换，直到找遍整个字符串。
        返回指向下一个标记串。当没有标记串时则返回空字符NULL。
  
  举例：
      // strtok.c
      
      #include <syslib.h>
      #include <string.h>
      #include <stdio.h>
      main()
      {
        char *s="Golden Global View";
        char *d=" ";
        char *p;
        
        clrscr();
        
        p=strtok(s,d);
        while(p)
        {
          printf("%s/n",s);
          strtok(NULL,d);
        }
        getchar();
        return 0;
      }*/