PATA1050 String Subtraction

1050 String Subtraction

Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking all the characters in S​2​​ from S​1​​. Your task is simply to calculate S​1​​−S​2​​ for any given strings. However, it might not be that simple to do it fast.

Input Specification:

Each input file contains one test case. Each case consists of two lines which gives S​1​​ and S​2​​, respectively. The string lengths of both strings are no more than 10​4​​. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.

Output Specification:

For each test case, print S​1​​−S​2​​ in one line.

Sample Input:

  

   

    

     

    
    

     

      They are students.
     
    
   

    

     

    
    

     

      aeiou
     
    
  
 

Sample Output:

Thy r stdnts.
 

  

   

    

     

    
    

     

      #include<cstdio>
     
    
   

    

     

    
    

     

      #include<iostream>
     
    
   

    

     

    
    

     

      #include<cstring>
     
    
   

    

     

    
    

     

      #include<algorithm>
     
    
   

    

     

    
    

     

      using namespace std;
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      int main()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     

      // freopen("C://坚果云//算法//String Subtractionin.txt","r",stdin);
     
    
   

    

     

    
    

     

      // freopen("C://坚果云//算法//String Subtractionout.txt","w",stdout);
     
    
   

    

     

    
    

     
	bool b[1000];
     
    
   

    

     

    
    

     

      	string a,c;
     
    
   

    

     

    
    

     
	getline(cin,a);
     
    
   

    

     

    
    

     
	getline(cin,c);
     
    
   

    

     

    
    

     
	for(int i=0;i<1000;i++)
     
    
   

    

     

    
    

     

      	{
     
    
   

    

     

    
    

     

      		b[i]=true;
     
    
   

    

     

    
    

     

      	}
     
    
   

    

     

    
    

     
	for(int j=0;j<c.length();j++)
     
    
   

    

     

    
    

     

      	{
     
    
   

    

     

    
    

     

      				b[c[j]]=false;
     
    
   

    

     

    
    

     

      	}
     
    
   

    

     

    
    

     
	for(int j=0;j<a.length();j++)
     
    
   

    

     

    
    

     

      	{
     
    
   

    

     

    
    

     
		if(b[a[j]]==true)
     
    
   

    

     

    
    

     

      		{
     
    
   

    

     

    
    

     
			printf("%c",a[j]);
     
    
   

    

     

    
    

     
			
     
    
   

    

     

    
    

     

      		}
     
    
   

    

     

    
    

     

      	}
     
    
   

    

     

    
    

     

      // fclose(stdin);
     
    
   

    

     

    
    

     

      // fclose(stdout);
     
    
   

    

     

    
    

     
	return 0;
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: blog.csdn.net，作者：沧夜2021，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/CANGYE0504/article/details/88836364