1050 String Subtraction （20 分）

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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，通过使用字符映射表，快速地从第一个字符串中移除第二个字符串中出现的所有字符。文章详细解释了算法的实现过程，并提供了完整的C++代码示例。

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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.

#include<bits/stdc++.h>
using namespace std;
int main()
{
string s1,s2;
int v[1005]={0};
getline(cin,s1);
getline(cin,s2);
for(int i=0;i<s1.length();i++)
{
v[s1[i]]=1;
}
for(int i=0;i<s2.length();i++)
{
if(v[s2[i]]==1)
v[s2[i]]=0;
}
for(int i=0;i<s1.length();i++)
{
if(v[s1[i]]>0)
cout<<s1[i];
}
}