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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.
Input Specification:
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.
Output Specification:
For each test case, print S~1~ - S~2~ in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.思路：
给出两个字符串，在第一个字符串中删除第二个字符串中出现过的所有字符并输出。
C++：
#include "cstring"
#include <cstdio>
#include "iostream"
using namespace std;
char s1[10001];
char s2[10001];
int main(){int len1,len2;cin.getline(s1,10001);cin.getline(s2,10001);len1=strlen(s1);len2=strlen(s2);bool flag[256]={false};for (int i=0;i<len2;i++){flag[s2[i]]=true;}for (int i=0;i<len1;i++){if (!flag[s1[i]]){printf("%c",s1[i]);}}return 0;
}