1050. String Subtraction (20)

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.

IDEA

1.题目中说所有的字符都在ASCII表中可见，计算速度要快。所以讲s2需要去掉的字符在ASKII中做标记。

然后遍历s1，判断每个字符在ASKII表中的标记，决定是否输出

2.ASKII范围是0-127 

CODE

#include<iostream>#include<cstring>using namespace std;int main(){string s1,s2;getline(cin,s1);getline(cin,s2);//cout<<s1<<endl<<s2;int flag[200]={0};//ASKII范围0-127 for(int i=0;i<s2.size();i++){flag[s2[i]]=1;}for(int i=0;i<s1.size();i++){if(!flag[s1[i]]){cout<<s1[i];}}return 0;}