pat 1050 String Subtraction(20 分)

1050 String Subtraction(20 分)

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.
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <map>
 6 #include <stack>
 7 #include <vector>
 8 #include <queue>
 9 #include <set>
10 #define LL long long
11 using namespace std;
12 const int MAX = 1e4 + 10;
13 
14 string s1, s2;
15 int book[MAX] = {0}, len1, len2;
16 
17 int main()
18 {
19 //    freopen("Date1.txt", "r", stdin);
20     getline(cin, s1);
21     getline(cin, s2);
22     len1 = s1.size(), len2 = s2.size();
23     for (int i = 0; i < len2; ++ i) book[s2[i]] = 1;
24     for (int i = 0; i < len1; ++ i)
25         if (!book[s1[i]])
26             printf("%c", s1[i]);
27     return 0;
28 }

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转载自www.cnblogs.com/GetcharZp/p/9589426.html