1050. String Subtraction (20)

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本文介绍了一种计算两个字符串差集的高效算法，并提供了完整的C++实现代码。输入为两个字符串S1和S2，输出为从S1中去除所有S2中存在的字符后的剩余字符串。文章通过使用标准模板库(map和string)来简化操作并提高效率。

Given two strings S₁ and S₂, S = S₁ - S₂ is defined to be the remaining string after taking all the characters in S₂from 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.

#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;

int main(){
    string s1,s2,output;
    getline(cin, s1);
    getline(cin, s2);
    map<char, bool> c_map;
    for(int i=0; i<256; i++)c_map.insert(make_pair( i , 1));
    for(string::iterator it=s2.begin() ; it!=s2.end() ; ++it)c_map[*it]=0;
    for(string::iterator it=s1.begin() ; it!=s1.end() ; ++it)if(c_map[*it])output+=*it;
    cout<<output;
    return 0;
}