Create 25-Cry-shuiyue-杨晶.md

Cry/submissions/Cry/submissions/25-Cry-shuiyue-杨晶.md

@@ -2,17 +2,95 @@
 >（开始的时候想直接在windows系统上进行配置，但是我的电脑无法开启虚拟机平台
 >>（开启进度(各种方式）永远卡在5.9%/14.6%/37.8%/56.2，直到失败）      
 >最后在别人的帮助下才装Arch liunx，然后再配置下载，直到这个星期六中午才配好。       
-！[结果如图](<img width="563" height="132" alt="4a029d11a14419121523e1df51db36bb" src="https://github.com/user-attachments/assets/9beff7a4-1876-4c9a-a9a5-b9fb64cf97e3"/>)     
+！[结果如图]<img width="563" height="132" alt="4a029d11a14419121523e1df51db36bb" src="https://github.com/user-attachments/assets/9beff7a4-1876-4c9a-a9a5-b9fb64cf97e3"/ >    
 ##RSA算法学习  
 1.RSA是一种非对称加密算法，依靠大素数分解困难，费马小定理与二次探测原理(即米勒拉宾素性测试)为基础。    
 2.先生成两位512位的素数*p,q*，然后通过**欧拉函数**计算出其小于其乘积N的素数的个数&phi(N)。   
 3.选取公钥e介于1 和 步骤二计算出的欧拉函数值之间，且gcd(e，&phi) = 1。 
 4.算私钥*d*满足e*d = 1(mod&phi(N))。
 >>
 >>进度过慢,我加几张笔记吧(假期里的老本)
->>！[笔记1](<img width="960" height="1280" alt="005d906623b8b3fa8efed11e72aef631_720" src="https://github.com/user-attachments/assets/ec2431ff-5eac-4bec-a0f1-af0a9f27e6c2"/>)
->>![笔记2](<img width="960" height="1280" alt="3b381265ec7524ac9b3fa4c2e52d487e_720" src="https://github.com/user-attachments/assets/14d45050-dc57-48f5-ae3f-e659c0c0049f"/>)      
->>![笔记3](<img width="960" height="1280" alt="73f2174de882794114424646e9e96678_720" src="https://github.com/user-attachments/assets/50ef4606-cd8c-4eb2-b324-3be38a12a3b2"/>)      
+>>！[笔记1]<img width="96" height="128" alt="005d906623b8b3fa8efed11e72aef631_720" src="https://github.com/user-attachments/assets/ec2431ff-5eac-4bec-a0f1-af0a9f27e6c2"/>
+>>![笔记2]<img width="96" height="128" alt="3b381265ec7524ac9b3fa4c2e52d487e_720" src="https://github.com/user-attachments/assets/14d45050-dc57-48f5-ae3f-e659c0c0049f"/>        
+>>![笔记3]<img width="96" height="128" alt="73f2174de882794114424646e9e96678_720" src="https://github.com/user-attachments/assets/50ef4606-cd8c-4eb2-b324-3be38a12a3b2"/>          
+
+##题目一：eeeeez_rsa
+题目：
+```python
+  from Crypto.Util.number import *
+  from secrets import randbits
+
+  p = getPrime(512)
+  q = getPrime(512)
+
+  n = p * q
+  e = 65537
+
+  m = bytes_to_long(flag)
+
+  c = pow(m, e, n)
+
+  print("p =", p)
+  print("q =", q)
+  print("e =", e)
+  print("c =", c)
+  '''     
+  p = 7833526559350210716763736624276871338973265302039384005037668270722459749111510212645578582715412039060908556429504391788904499303021963918033838457712021
+  q = 11274522482114648167653425707657117942167215969612324249057858493398092816663094339833695146914712716091767834292452373966791325922322967827879446489910963
+  e = 65537
+   c=1022172159188954584457305025459579586118234885650614560548034824653447703208511380139038324539142727422462790049010578748622777418895745000669136397439695287885078  
+'''
+```
+
+
+
+解答：<img width="667" height="356" alt="d9fe777716c9b49a3ba4ed4f22e505aa" src="https://github.com/user-attachments/assets/f57d9715-9273-425a-8099-b81cf1ae9896" />            
+解题思路：     
+1.知识了解：Crypto.Util.number 是 PyCryptodome 库中的一个工具模块，专门处理大整数的密码学运算 
+2.题目给出了n分出的两个素数p,q,给出了私钥e,密文c,明文m。      
+3.在题目中发现m是 flag 转化成的数字，那么要得到flag,就要先把m 转化成字节串，再转化成字符串，于是    
+先'long_to_bites',然后再decode,就得到了flag。     
+
+
+题目二：What is dpdq         
+题目：
+```python
+from Crypto.Util.number import *
+
+p = getPrime(512)
+q = getPrime(512)
+
+n = p * q
+
+m = bytes_to_long(flag)
+
+c = pow(m, e, n)
+
+phi = (p - 1) * (q - 1)
+d = inverse(e, phi)
+
+dp = d % (p - 1)
+dq = d % (q - 1)
+
+print("p =", p)
+print("q =", q)
+print("dp =", dp)
+print("dq =", dq)
+print("c =", c)
+'''
+p = 7815414185491141116816659592648655093824828684096724566017773298150863675227130435782645950134326106977982747903113904523205447790009729530724322503221833
+q = 9989016009119164140172559452397593815416653032520645020530362343604703571290641132626496035194681031089349635895258123948622879868985300264027551088746747
+dp = 4010681140217557380422935065992638785960856284290416720391683347779267392850433355451759290094414691393922757792964689212573746434435619933431808206484225
+dq = 2650091114188243387783699150080671432452650354401886796343461099352350290584408461982954461814393687345324352950407449243857759926179828428073409383707519
+c = 49442655923625309909385017545371837339079019962533687954830521594031788193892512818961352203881969632046490177930759665200677302956816647378151364853230118743634390053041583940566732297636464662866406886764229490729837423276227228984024967061624432225315562131278455141688731687433597958272776740673914705069
+'''
+```     
+题目解答：    <img width="1374" height="288" alt="35789eaca7549618262df95ecf639364" src="https://github.com/user-attachments/assets/c54b713e-495d-434b-8d34-bd27bcb22e6f" />
+题目分析：      
+1.题目相较于第一题，给出了dp,dq,dp与dq是p,q减1后去模的结果，可以减少计算。     
+2.通过dp,dq先解出密钥的一部分，然后利用中国剩余定理解得 m。     
+3.将m 转换为字符型，得到flag。     
+
 
 ##为什么要选择这个方向：  
 >实践出真知