破解密码之破解数字+字母密码篇----java实现

最新推荐文章于 2025-06-30 09:37:12 发布

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最新推荐文章于 2025-06-30 09:37:12 发布 · 1.3w 阅读

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#java

这篇博客介绍了如何用Java实现一个程序，通过穷举法来破解由数字和字母组成的密码。程序逐级生成并检查从一位到六位的所有可能组合，直到找到与目标密码匹配的字符串。

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import java.lang.*;
public class PojiePassword{
   static String real="ZYXLXP";//真实密码
   static String pass="";//执行循环操作找出来的与真实密码相等的字符串
   static String prod="";//中间产生的字符串
   static  String [] s={"0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
//定义一个有62个字符串的字符串数组
//生成一个指定位数的密码，每一位有以上62种可能,n=6时，第一个为000000，最后一个为ZZZZZZ,共//62^6种可能。n=7时，第一个为0000000，最后一个为ZZZZZZZ，共62^7种可 能。n=8时，第一个为//00000000，最后一个为ZZZZZZZZ，共62^8种可能
//62^6种情况，累死也列不出来，，，


public static void main(String []args)   throws InterruptedException{


//循环遍历出数组中的元素，列出数组中的字符可以组成的所有一位字符串，共62^1种可能
   
       for(int i=0;i<62;i++){
         prod=s[i];
         if(prod.equals(real)){
             pass=prod;
             System.out.println("执行"+(i+1)+"次操作,找到真实密码，为"+pass); 
             break;
         }