算法分析

模拟

• 1、将原来的密码pass1通过转换变成pass2，若pass1和pass2不一样，则表示该密码已被修改，待输出
• 2、若没有密码被修改过，还需要注意数量是1个还是多个
• 3、若有密码被修改过，则把修改过的密码按顺序输出

时间复杂度 $O(10 * n)$

参考文献

pat

Java 代码

import java.util.Scanner;

public class Main {
    static int N = 1010;
    static String[] ans = new String[N];
    static int k = 0;
    static String change(String s)
    {
        char[] temp = s.toCharArray();
        for(int i = 0;i < temp.length;i ++)
        {
            char t = s.charAt(i);
            if(t == '1') t = '@';
            else if(t == '0') t = '%';
            else if(t == 'l') t = 'L';
            else if(t == 'O') t = 'o';
            temp[i] = t;
        }
        return String.valueOf(temp);
    }
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int res = 0;
        for(int i = 0;i < n;i ++)
        {
            String id = scan.next();
            String pass1 = scan.next();

            String pass2 = change(pass1);//变换后的密码
            if(!pass1.equals(pass2)) 
            {
                res ++;
                ans[k ++] = id + " " + pass2;
            }
        }
        if(res == 0) 
        {
            int t = n;
            System.out.println("There " + (t == 1 ? "is" : "are") + " " + t + " account" + (t == 1 ? "" : "s") + " and no account is modified");
        }
        else
        {
            System.out.println(res);
            for(int i = 0;i < k;i ++) System.out.println(ans[i]);
        }
    }
}