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HDU1228 A + B

A + B


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11043    Accepted Submission(s): 6376


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

#include<stdio.h>
#include<string.h>
char str[100];
char s[100][6];
char flag[11][6]={"zero","one","two","three","four","five","six","seven","eight","nine","ten"};
int main()
{
    int i,j,k,u,n,m,v1,v2,num;
    while(gets(str))
    {
        n=strlen(str);
        k=0;u=0;
        for(i=0;i<n;i++)
        {
           if(str[i]!=' ')
           s[k][u++]=str[i];
           else
           {
               k++;u=0;
           }
        }
        if(strcmp(s[0],"zero")==0&&strcmp(s[2],"zero")==0)
        {break;}
        else
        {
            v1=0;v2=0;num=1;
           for(i=0;i<k;i++)
           {
               if(strcmp(s[i],"+")!=0)
               {
                  for(j=0;j<10;j++)
                  {
                     if(strcmp(s[i],flag[j])==0)
                     {
                           if(num==1)
                           v1=v1*10+j;
                           else
                           v2=v2*10+j;
                     }
                  }
               }
               else
               {
                  num=2;
               }
           }
           //printf("%d+%d=",v1,v2);
           printf("%d\n",v1+v2);
           memset(str,0,sizeof(str)); 
           memset(s,0,sizeof(s)); 
        }             
    }
    return 0;
}









最后更新：2017-04-03 12:54:47

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