A1050. 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.

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
/*
A1050 给出两个字符串，在第一个字符串中删去第二个字符串中出现过的所有字符并输出
*/
int main()
{
    char str1[10010];
    char str2[10010];
    gets(str1);
    gets(str2);
//    char str1[10010]="They are students.";
//    char str2[10010]="aeiou";
    int str1Length=strlen(str1);
    int str2Length=strlen(str2);
    bool hashTable[256]={false};
    for(int i=0;i<str2Length;++i){
        char c2=str2[i];//临时存放字符串2的字符
        if(hashTable[c2]==false){
            hashTable[c2]=true;
        }
    }
    for(int j=0;j<str1Length;++j){//注意这里长度不要写成str2Length,容易惯性思维犯错
        char c1=str1[j];//临时存放字符串1的字符
        if(hashTable[c1]==true){
            continue;
        }
        else{
            printf("%c",c1);
        }
    }
    return 0;
}