HDU 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

39096

#include<string.h>char num[10][10] = {"zero","one","two","three","four","five","six","seven","eight","nine"};int f(char s[]){int i;for(i=0;i<10;i++)  //将单个英文单词转换成数字 {if(!strcmp(num[i],s)) break;}return i;} int main(void){int a,b;char str[100];while(1){a=b=0;while(scanf("%s",str)&&strcmp(str,"+")){a=a*10+f(str);}while(scanf("%s",str)&&strcmp(str,"=")){b=b*10+f(str);}if(a==0&&b==0) break;//或者return 0;printf("%d\n",a+b);}return 0;}