hdoj-1228-A+B

Problem Description

读入两个小于100的正整数A和B,计算A+B.
需要注意的是:A和B的每一位数字由对应的英文单词给出.

 

Input

测试输入包含若干测试用例,每个测试用例占一行,格式为"A + B =",相邻两字符串有一个空格间隔.当A和B同时为0时输入结束,相应的结果不要输出.

 

Output

对每个测试用例输出1行,即A+B的值.

 

Sample Input

   
   
   
   

    
    
    
    one + two =
three four + five six =
zero seven + eight nine =
zero + zero =
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
90
96
   
   
   
   

中文题面，意思很好懂。

直接模拟过去就好了

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
    int main()  
    {  
        char str[30],s1[20];  
        int i,j,len,s,sum1,sum2,flag;  
        while(gets(str)!=NULL)  
        {  
            len=strlen(str);  
            sum1=sum2=0;   
            flag=0;   
            for(i=0,j=0;i<len;i++)  
            {  
                if(str[i]>='a'&&str[i]<='z')   
                    s1[j++]=str[i];  
                else if(str[i]==' '&&str[i-1]!='+')  
                {  
                    s1[j]='\0'; 
                    if(!strcmp(s1,"zero")) s=0;   
                    else if(!strcmp(s1,"one")) s=1;  
                    else if(!strcmp(s1,"two")) s=2;  
                    else if(!strcmp(s1,"three")) s=3;  
                    else if(!strcmp(s1,"four")) s=4;  
                    else if(!strcmp(s1,"five")) s=5;  
                    else if(!strcmp(s1,"six")) s=6;  
                    else if(!strcmp(s1,"seven")) s=7;  
                    else if(!strcmp(s1,"eight")) s=8;  
                    else if(!strcmp(s1,"nine")) s=9;  
                    j=0;   
                    if(!flag)  
                        sum1=sum1*10+s;  
                    else   
                        sum2=sum2*10+s;   
                }  
                else if(str[i]=='+')  
                    flag=1;   
            }  
            if(sum1==0&&sum2==0)  
                break;   
            else  
                printf("%d\n",sum1+sum2);  
        }  
        return 0;  
    }