hdu 1228 A+B

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A + B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10803 Accepted Submission(s): 6228

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
已AC（程序如下）：
#include<stdio.h>#include<string.h>char a[10][6]={"zero","one","two","three","four","five","six","seven","eight","nine"};int search(char b[]){int i;for(i=0;i<10;i++){if(strcmp(b,a[i])==0)break;}return i;}int main(){char c[10];int m,n;while(1){m=0;while(scanf("%s",c)&&strcmp(c,"+")!=0)m=m*10+search(c);n=0;while(scanf("%s",c)&&strcmp(c,"=")!=0)n=n*10+search(c);if(m==0&&n==0)return 0;elseprintf("%d\n",m+n);}return 0;}