1050 String Subtraction (20 分)

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 1. 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.

注意：考场上可能关闭快的服务器，只留慢的，所以两层for循环不保险
两层for循环：

#include<bits/stdc++.h>
using namespace std;
const int maxn=10010;

int main(){
    char s1[maxn];
    char s2[maxn];
    cin.getline(s1,maxn);
    cin.getline(s2,maxn);
    for(int i=0;s1[i]!='';i++){
        for(int j=0;s2[j]!='';j++){
            if(s1[i]==s2[j]){
                s1[i]='0';
            }
        }
    }
    for(int i=0;s1[i]!='';i++){
        if(s1[i]!='0'){
            printf("%c",s1[i]);
        }
        else{
            continue;
        }
    }
    printf("
");
    return 0;
}

优化代码:

#include<bits/stdc++.h>
using namespace std;
const int maxn=100010;
char s1[maxn];
char s2[maxn];
bool flag[maxn]={false};
int main(){
    cin.getline(s1,maxn);
    cin.getline(s2,maxn);
    int ls1=strlen(s1);
    int ls2=strlen(s2);
    for(int i=0;i<ls2;i++){
        flag[s2[i]]=true;//将字符转化为整形，作为下标存储
    }
    for(int i=0;i<ls1;i++){
        if(flag[s1[i]]==false){//相同下标下为false，则输出
            printf("%c",s1[i]);
        }
    }
    printf("
");
    return 0;
}