Given two strings S?1 and S2, S=S1?S2 is defined to be the remaining string after taking all the characters in S?2 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 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.
Output Specification:
For each test case, print S1?S2in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.
题意：输入两个字符串，删掉第一个字符串中在第二个字符串中出现的字符
思路：用一个长度为256的数组纪录第二个字符串中出现的字符，一个坑点就是数组长度必须最好设置成100010这么大，刚设置成10010的话只有14分，看了柳神的博客才改过来，题目要求是10的4次方就行了。。但是这个地方可能是测试系统的原因，要求长度大一点。
#include<iostream>
#include<cstdio>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>using namespace std;bool flag[256] = {
false};char s1[100100];
char s2[100100];int main() {
cin.getline(s1,100000);cin.getline(s2,100000);int length1 = strlen(s1);int length2 = strlen(s2);for(int i = 0;i < length2;i++) {
flag[s2[i]] = true;}for(int i = 0;i < length1;i++) {
if(!flag[s1[i]]) {
cout << s1[i];}}return 0;
}