PAT 1050 String Subtraction

最新推荐文章于 2024-05-17 06:07:48 发布

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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，通过标记已知字符串中的字符，从而快速找出目标字符串中未出现过的字符。文章提供了一个示例代码，展示了如何使用C++实现这一算法。

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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.

题意:输出S1中在S2中未出现过的字符

思路:标记S2中出现过的字符

代码

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    string S1,S2;
    getline(cin,S1);
    getline(cin,S2);
    int Mark[200],Len;
    memset(Mark,0,sizeof(Mark));
    Len=S2.size();
    for(int i=0;i<Len;i++)
    {
        int x=S2[i];
        Mark[x]=1;
    }
    Len=S1.size();
    for(int i=0;i<Len;i++)
    {
        int x=S1[i];
        if(!Mark[x])
            cout<<S1[i];
    }
    return 0;
}