PAT 1050 String Subtraction [hash]

本文主要是介绍PAT 1050 String Subtraction [hash]，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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.

----------------------------------这是题目和解题的分割线----------------------------------

However, it might not be that simple to do it fast. 题目已经给出明示不能用两层循环啦。

#include<stdio.h>
#include<string.h>int main()
{int exsit[128] = {},i;char x1[10005],x2[10005];gets(x1);gets(x2); //ASCII码表的起始值是'\0'  for(i=0;i<strlen(x1);i++)exsit[x1[i]-'\0'] = 1;for(i=0;i<strlen(x2);i++)exsit[x2[i]-'\0'] = 0;for(i=0;i<strlen(x1);i++)if(exsit[x1[i]-'\0']) printf("%c",x1[i]);
}

看书上才知道，原来bool数组的下标除了是可以int外还可以是char类型。在这种只需判断有无的题中会比较方便。

#include<cstdio>
#include<iostream>
#include<string>using namespace std;int main()
{bool exsit[128];int i;string x1,x2;getline(cin,x1);getline(cin,x2); for(i=0;i<x1.length();i++)exsit[x1[i]] = true;for(i=0;i<x2.length();i++)exsit[x2[i]] = false;for(i=0;i<x1.length();i++)if(exsit[x1[i]]) printf("%c",x1[i]);
}

 

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