hdu 1228 A + B -----模拟

A + B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7852    Accepted Submission(s): 4417

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

                 直接 把字符串转化为数字。  遇到 '+'   换数字。  遇到  ‘=’  停止输入。

//Memory: 216 KB		Time: 0 MS
//Language: C++		Result: Accepted
#include<stdio.h>
#include<string.h>
int main()
{
	char a[6][10];
	int i,b[2],j,temp,ten,c;
	while(scanf("%s",a[0])!=EOF)
	{
		ten=1;
		c=0;
		b[0]=b[1]=0;
		for(i=0;;i++)
		{
			if(i!=0)
			scanf("%s",a[i]);
			if(a[i][0]=='=') break;
			if(strcmp(a[i],"zero")==0)
				temp=0;
			if(strcmp(a[i],"one")==0)
				temp=1;
			if(strcmp(a[i],"two")==0)
				temp=2;
			if(strcmp(a[i],"three")==0)
				temp=3;
			if(strcmp(a[i],"four")==0)
				temp=4;
			if(strcmp(a[i],"five")==0)
				temp=5;
			if(strcmp(a[i],"six")==0)
				temp=6;
			if(strcmp(a[i],"seven")==0)
				temp=7;
			if(strcmp(a[i],"eight")==0)
				temp=8;
			if(strcmp(a[i],"nine")==0)
				temp=9;
			if(a[i][0]=='+')
			{
				ten=1;
				c=1;
			}
			else
			{
		        b[c]=b[c]*ten+temp;
			     ten*=10;
			}
		}
		if(b[0]+b[1]==0) break;
		printf("%d\n",b[0]+b[1]);
	}
	return 0;
}