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 10e4. 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.

题目大意：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～
分析：用数组变量标记str2出现过的字符为，输出str1的时候检查是否被标记过。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
using namespace std;int main(void)
{#ifdef testfreopen("in.txt","r",stdin);//freopen("in.txt","w",stdout);clock_t start=clock();#endif //testchar s1[10010],s2[10010];fgets(s1,10010,stdin);//PAT不支持gets，要用fgets代替fgets(s2,10010,stdin);int num[300]={0};for(int i=0;s2[i];++i)num[s2[i]]=1;for(int i=0;s1[i];++i)if(num[s1[i]]==0)printf("%c",s1[i]);printf("\n");#ifdef testclockid_t end=clock();double endtime=(double)(end-start)/CLOCKS_PER_SEC;printf("\n\n\n\n\n");cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位#endif //testreturn 0;
}