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88 changes: 84 additions & 4 deletions Cry/submissions/Cry/submissions/25-Cry-shuiyue-杨晶.md
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>（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="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"/>

##题目一
1. eeeeez_rsa

```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 = 10221721591889545844573050254595795861182348856506145605480348246534477032085113801390383245391427274224627900490105787486227774188957450006691363974396952878850785124275814042623413354128640305152567667226654321244510669777870256213487113315863197287562256336752636572876416053482569293397021584043886350655
'''
```
###题目解答
<img width="667" height="356" alt="d9fe777716c9b49a3ba4ed4f22e505aa" src="https://github.com/user-attachments/assets/8835b832-146e-4da2-8566-847833e1becf" />

###题目分析
1.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.发现题目中 m 是明文的数字形式，于是想先得到m。
3.通过'pow'得到m,再通过'long_to_bytes'转化m 得到字节型，再'decode'得到flag。

##题目二
2. What is dpdq

```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/f98abe7e-8035-4c97-836d-2117bbe617a2" />

###题目解析
1.相比第一题，这个题目引入了dp.dq，简化了运行时的运算。
2.知识补充（自我提醒）：
>题目中的dp,dq是由p,q分别减一再取模得到，可以简化计算。
3.由于dp,dq可看作密钥的一部分，故而先'pow'算出dp,dq。
4.利用中国剩余定理得到m，再将m转化为字符型，得到flag。
##为什么要选择这个方向：
>实践出真知

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