HDOJ 1228 A + B

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

   
   
   
   

    
    
    
    3
90
96
   
   
   
   

 

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;
}