HDOJ 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): 13081 Accepted Submission(s): 7705

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

题目比较典型，尤其是在控制输入方面，用字符串数组配合strcmp控制输入做法很灵活。

具体代码如下：

#include<stdio.h>#include<string.h>char str[10][6]={"zero","one","two","three","four","five","six","seven","eight","nine"};int search(char s[]){int i;for(i=0;i<10;i++){if(strcmp(s,str[i])==0)   return i;}}int main(){char s[20];int a,b;while(1){a=0;while(scanf("%s",s)&&strcmp(s,"+")!=0)a=a*10+search(s);b=0;while(scanf("%s",s)&&strcmp(s,"=")!=0)b=b*10+search(s);if(a==0&b==0)   return 0;else   printf("%d\n",a+b);}return 0;}

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