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;}