杭电1228 A+B（英文版）

A + B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12153    Accepted Submission(s): 7103

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

 

CODE：

#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
	int n;
	char a[5000];
	char cha[20][20];
	while(gets(a))
	{
		
		int i,j,k,len;
		len=strlen(a);
		k=0;
		for(i=0;i<len;)
		{
			for(j=0;;j++)
			{
				if(a[i]!=' ')
				{
					cha[k][j]=a[i];
					i++;
				}
				else
				{
					k++;i++;break;
				}
			}
		}
		
		int t=0,b[1000];
		for(i=0;i<k;i++)
		{
			if(!strcmp(cha[i],"one"))
			b[t++]=1;
			else if(!strcmp(cha[i],"two"))
			b[t++]=2;
			else if(!strcmp(cha[i],"three"))
			b[t++]=3;
			else if(!strcmp(cha[i],"four"))
			b[t++]=4;
			else if(!strcmp(cha[i],"five"))
			b[t++]=5;
			else if(!strcmp(cha[i],"six"))
			b[t++]=6;
			else if(!strcmp(cha[i],"seven"))
			b[t++]=7;
			else if(!strcmp(cha[i],"eight"))
			b[t++]=8;
			else if(!strcmp(cha[i],"nine"))
			b[t++]=9;
			else if(!strcmp(cha[i],"zero"))
			b[t++]=0;
			else if(!strcmp(cha[i],"+"))
			b[t++]=-1;
			else
			continue;
		}
		//printf("%d@",b[1]);
		int temp;
		for(i=0;i<t;i++)
		if(b[i]==-1)
		{
			temp=i;break;
		}
		//printf("%d!",temp);
		int A,B;
		A=B=0;
		for(i=0;i<temp;)
		if(b[i]==0)
		i++;
		else break;
		for(;i<temp;i++)
		if(temp-i==2)
		A+=10*b[i];
		else
		A+=b[i];
		//printf("%d&",A);
		for(i=temp+1;i<t;)
		if(b[i]==0)
		i++;
		else break;
		for(;i<t;i++)
		if(t-i==2)
		B+=10*b[i];
		else
		B+=b[i];
		//printf("%d~",B);
		if(!(A||B))
		break;
		printf("%d\n",A+B);
		memset(a,'\0',sizeof(a));
		memset(cha,'\0',sizeof(cha));
		
	}
	return 0;
}