HDUOJ A+B

A + B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10595    Accepted Submission(s): 6104

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
  
  

 

分析：该题是英文的A+B求和，个人思路是用gets函数输入，然后在切割字符串求两数，最后求和：

AC代码：

#include <stdio.h>
#include <string.h>
#define N 100
int main()
{
	char s[N],str[N],num[10][8]={"zero","one","two","three","four","five","six","seven","eight","nine"};
	int a,b,k,i,j,len,flag,n[5];
	while(gets(s))
	{
		k=1;a=b=0;
		for(i=len=0;s[i]!='=';i++)
		{
			if(s[i]==' ' && len!=0)
			{
				str[len]='\0';
				//printf("%s\n",str);
				for(j=0;j<10;j++)
					if(strcmp(str,num[j])==0) break;
				n[k++]=j;
				len=0;
				if(s[i+1]=='+') flag=k;
			}
			else if(s[i]>='a' && s[i]<='z')
			{
				str[len++]=s[i];
			}
		}
			if(flag==2)
			{
				a=n[1];
				if(k==3)
					b=n[2];
				else
					b=n[2]*10+n[3];
			}
			else if(flag=3)
			{
				a=n[1]*10+n[2];
				if(k==4)
					b=n[3];
				else
					b=n[3]*10+n[4];
			}
		//printf("%d %d %d %d,a=%d,b=%d\n",n[1],n[2],n[3],n[4],a,b);
		if(a==0 && b==0) break;
		else
			printf("%d\n",a+b);
	}
	return 0;
}

下面为网上的其他代码,

代码1：

#include <stdio.h>   
#include <stdlib.h>   
#include <string.h>   
  
int strToInt(char *);  
  
int main()  
{  
    char str[6];  
    int a, b, c;  
  
    while(scanf("%s",str))  
    {  
        a = strToInt(str);  
        while(scanf("%s",str))  
        {  
            if(str[0] != '+' && str[0] != '=')  
            {  
                b = strToInt(str);  
                a = a * 10 + b;  
            }else if(str[0] == '+')  
            {  
                c = a;  
                a = 0;  
            }else if(str[0] == '=')  
            {  
                break;  
            }  
        }  
  
        if(!a && !c)  
        {  
            break;  
        }else  
        {  
            printf("%d\n", a + c);  
        }  
    }  
  
    return 0;  
}  
  
int strToInt(char *str)  
{  
    int num;  
  
    if(strcmp(str, "zero") == 0)  
    {  
        num = 0;  
    }else if(strcmp(str, "one") == 0)  
    {  
        num = 1;  
    }else if(strcmp(str, "two") == 0)  
    {  
        num = 2;  
    }else if(strcmp(str, "three") == 0)  
    {  
        num = 3;  
    }else if(strcmp(str, "four") == 0)  
    {  
        num = 4;  
    }else if(strcmp(str, "five") == 0)  
    {  
        num = 5;  
    }else if(strcmp(str, "six") == 0)  
    {  
        num = 6;  
    }else if(strcmp(str, "seven") == 0)  
    {  
        num = 7;  
    }else if(strcmp(str, "eight") == 0)  
    {  
        num = 8;  
    }else if(strcmp(str, "nine") == 0)  
    {  
        num = 9;  
    }  
      
    return num;  
}  

该方法是处理含空格字符串的一种处理方法，过程是用scanf函数分段输入处理，很值得学习和借鉴。

未完待续...