求助，有哪位大牛可以帮我写一下这道题，谢谢。。。

[SmallKind]

用英文单词模拟数学计算
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。
具体的输入输出格式规定如下：
输入格式：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zero时输入结束，相应的结果不要输出。
输出格式：对每个测试用例输出1行，即A+B的值。

输入样例：

one + two =
three four + five six =
zero seven + eight nine =
zero + zero =

输出样例：
three
nine zero
nine six

[W170532934]

自己读取字符串然后进行解析吧。

[SmallKind]

[Quote=引用 1 楼 的回复:]
自己读取字符串然后进行解析吧。
[/Quote]
就是那个最后求出结果的不是数字，就是在把他转换成字符怎么写，就是这个不知道写啊，比如one + two =3，然后在把3转换成three，这个不知道啦，哎，很纠结，麻烦啊

[DataChat.Club]

one + two =3是咋算的？

反过来就行了吧

[Quote=引用 2 楼 的回复:]

引用 1 楼 的回复:
自己读取字符串然后进行解析吧。

就是那个最后求出结果的不是数字，就是在把他转换成字符怎么写，就是这个不知道写啊，比如one + two =3，然后在把3转换成three，这个不知道啦，哎，很纠结，麻烦啊
[/Quote]

[代码Dog]

你把每个数字字符串与具体的数字映射起来，计算結果反映射输出就行，不用字符串解析这么麻烦。。

[SmallKind]

汗，我在看看。。谢谢

[DataChat.Club]

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//字符串转换成数字
int str2int(char a[10])
{
	char c, d;
	int num;
	c = a[0];
	d = a[1];
	switch (c)
	{
	case 'z': num = 0;break;
	case 'o': num = 1;break;
	case 't': switch(d)
			  {
			  case 'w': num = 2;break;
			  case 'h': num = 3;break;
			  }break;
	case 'f':switch(d)
			 {
			 case 'o':num = 4;break;
			 case 'i':num = 5;break;
			 }break;
	case 's':switch(d)
			 {
			 case 'i':num = 6;break;
			 case 'e':num = 7;break;
			 }break;
	case 'e':num = 8;break;
	case 'n':num = 9;break;
	}
	return num;
}

//数字转换成字符串
char *int2str(char n)
{	
	switch(n)
	{
	case '1': return "one ";break;
	case '2': return "two ";break;
	case '3': return "three ";break;
	case '4': return "four ";break;
	case '5': return "five ";break;
	case '6': return "six ";break;
	case '7': return "seven ";break;
	case '8': return "eight ";break;
	case '9': return "nine ";break;
	case '0': return "zero ";break;
	default: return "\n";
	}
}

int main()
{
	FILE *fp, *out;
	char temp[6];
	char *p;
	int num1, num2;

	if ((fp = fopen("in.txt", "r")) == NULL)//in.txt放置原始输入数据
	{
		printf("cannot open file !\n");
		exit(0);
	}
	if ((out = fopen("temp1.txt", "w")) == NULL)//temp1.txt放置转换成数字后的数据
	{
		printf("cannot open file !\n");
		exit(0);
	}

	while (!feof(fp))
	{
		fscanf(fp, "%s", temp);
		if (strcmp(temp, "+") && strcmp(temp, "="))
		{
			fprintf(out, "%d", str2int(temp));
		}
		else
		{
			fprintf(out, "\n", 1);
		}	
	}
	fclose(fp);
	fclose(out);

	if ((fp = fopen("temp1.txt", "r")) == NULL)
	{
		printf("cannot open file !\n");
		exit(0);
	}
	if ((out = fopen("temp2.txt", "w")) == NULL)//temp2.txt放置计算后的数字结果
	{
		printf("cannot open file !\n");
		exit(0);
	}
	while (!feof(fp))
	{
		fscanf(fp, "%d\n%d", &num1, &num2);
		if (!num1 || !num2)
		{
			break;
		}
		else
		{
			fprintf(out, "%d\n", num1+num2);	
		}
	}
	fclose(fp);
	fclose(out);

	if ((fp = fopen("temp2.txt", "r")) == NULL)
	{
		printf("cannot open file !\n");
		exit(0);
	}
	if ((out = fopen("out.txt", "w")) == NULL)//out.txt放置最后结果
	{
		printf("cannot open file !\n");
		exit(0);
	}
	while (!feof(fp))
	{
		p = int2str(fgetc(fp));
		
		fprintf(out, "%s", p);
	}
	fclose(fp);
	fclose(out);
	return 0;
}

[hen_hao_ji]

这题是hdu 1228吧

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char strr[10][16]={"zero","one","two","three","four","five","six","seven","eight","nine"};
char str[16];
int Find(char st[])
{
    int i;
    for(i=0;i<=9;i++)
    if(strcmp(st,strr[i])==0)
    return i;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int i1,i2,a,b;
    while(1)
    {
        a=b=0;
        scanf("%s",str);
        i1=Find(str);
        a=i1;
        scanf("%s",str);
        while(strcmp(str,"+")!=0)
        {
            i2=Find(str);
            a=i2+i1*10;
            scanf("%s",str);
        }
        //printf("%d\n",a);
        scanf("%s",str);
        i1=Find(str);
        b=i1;
        scanf("%s",str);
        while(strcmp(str,"=")!=0)
        {
            i2=Find(str);
            b=i2+i1*10;
            scanf("%s",str);
        }
        //printf("%d\n",b);
        if(a+b==0)
        break;
        else
        printf("%d\n",a+b);
    }
    return 0;
}

[sxldfang]

也来一个，看看咋样~~~

#include<stdio.h>
char *p[]={"zero","one","two","three","four","five","six","seven","eight","nine","+","="};
void outInt(unsigned int n)
{
	if(n>=10)outInt(n/10);
	printf("%s ",p[n%10]);
}

int main()
{
	FILE *f=fopen("D:\\1.txt","r");
	char buff[10];
	int n1,n2,n3,m,i;
	while(!feof(f))
	{
		n1=n2=m=0;
		while(1)
		{
			fscanf(f," %s",buff);
			for(i=0;i<12;++i)
			{
				if(strcmp(buff,p[i])==0)
				{
					if(i==11)goto Calc;
					if(i==10)m=1;
					else if(m)n2=n2*10+i;
					else n1=n1*10+i;
					break;
				}
			}
		}
Calc:
		n3=n1+n2;
		if(n3!=0)outInt(n3);
		printf("\n");
	}
	fclose(f);
	return 0;
}

D:\1.txt 的内容：

one + two =  
three four + five six =  
zero seven + eight nine =  
zero + zero = 

屏幕输出：

three
nine zero
nine six


请按任意键继续. . .

[SmallKind]

谢谢各位大牛啦。。