标签：task open map ace sci one bsp -o fast
Given two strings S?1?? and S?2??, S=S?1???S?2?? is defined to be the remaining string after taking all the characters in S?2?? from S?1??. Your task is simply to calculate S?1???S?2?? for any given strings. However, it might not be that simple to do it fast.
Each input file contains one test case. Each case consists of two lines which gives S?1?? and S?2??, respectively. The string lengths of both strings are no more than 10?4??. 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.
For each test case, print S?1???S?2?? in one line.
They are students.
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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 }
pat 1050 String Subtraction（20 分）
标签：task open map ace sci one bsp -o fast
原文地址：https://www.cnblogs.com/GetcharZp/p/9589426.html
