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.
解题思路
hash思想, 将s2中的字符串对应asc码表示到vis中,访问过表示1 循环一遍对应s2,如果s2中的字符标记为访问过的那么就不输出。
解题代码
#include <iostream>
using namespace std;
int vis[10010];
int main(){
string s1, s2;
getline(cin, s1);
getline(cin, s2);
for (int i = 0; i < s2.size(); i++) vis[s2[i]] = 1;
for (int i = 0; i < s1.size(); i++){
if (vis[s1[i]]) continue;
cout << s1[i];
}
return 0;
}