【PAT】1050 String Substraction

最新推荐文章于 2023-01-29 11:02:29 发布

realxuejin

最新推荐文章于 2023-01-29 11:02:29 发布

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本文介绍了一种高效计算字符串差集的方法，通过避免使用暴力搜索来提高算法效率。详细阐述了实现步骤，并提供了示例代码，展示了如何通过标记字符是否存在来快速计算字符串差集。

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题目：http://pat.zju.edu.cn/contests/pat-101-103-1-2013-03-10/A
题目描述：

Given two strings S₁ and S₂, S = S₁ - S₂ is defined to be the remaining string after taking all the characters in S₂ from S₁. Your task is simply to calculate S₁ - S₂ 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 S₁ and S₂, respectively. The string lengths of both strings are no more than 10⁴. 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 S₁ - S₂ in one line.

Sample Input:

They are students.
aeiou

Sample Output:

Thy r stdnts.

分析：怎么样快速的查询时关键，使用暴搜一定会超时的。

以下代码来自于：http://blog.youkuaiyun.com/sunbaigui/article/details/8656788

    #include<iostream>  
    #include <set>  
    #include <vector>  
    #include <map>  
    #include <queue>  
    #include <stack>  
    #include <string>  
    #include <string.h>  
    #include <algorithm>  
    using namespace std;  
      
    #define  MAX 1000  
    int main()  
    {  
        string a, b;
        int i;
        getline(cin, a);  
        getline(cin, b);  
        vector<bool> existed(MAX, false);  
        for (i = 0; i < b.size(); ++i)  
            existed[b[i]] = true;  
        for ( i = 0; i < a.size(); ++i)  
            if (!existed[a[i]]) printf("%c",a[i]);  
        cout<<endl;
        return 0;  
    }

不知道各位还有没有更好的做法？