当前位置：首页 > news >正文

西安做网站哪家好/中国十大电商平台排名

news 2025/8/6 6:31:03

西安做网站哪家好,中国十大电商平台排名,成都培训机构排名前十,p2p网站建设石家庄复现比赛系列；希望对大家学习密码学有所帮助 crypto1_中秋月题目描述： fsskryenvkm~jl{ejs}jwflzsnpgmifq{{j{|suhzrjppnx|qvixt~whu hint: 经此古典密码加密后，密文还是大写该古典密码的密钥形式： keywordplaintext (plainte…

复现比赛系列；希望对大家学习密码学有所帮助

crypto1_中秋月

题目描述：

fsskryenvkm~jl{ejs}jwflzsnpgmifq{{j{|suhzrjppnx|qvixt~whu

hint:

经此古典密码加密后，密文还是大写

该古典密码的密钥形式：
keyword+plaintext (+plaintext…+plaintext)

自动钥匙⊕

显然是自动密钥加密没错了

但是需要注意到这里给出的密文并不是这种加密得到的密文（应该是大写的）

可能经过了某种变换；

题目中还有⊕；可能是异或

使用cyberchief试一试异或（cyberchief的异或选项有爆破功能）

得到一种情况恰好全部为大写字母：

YLLTMFZQITRAUSDZULBUHYSELQOXRVYNDDUDCLJWEMUOOQGCNIVGKAHWJ5

github上搜到自动密钥解密的脚本break_autokeyGitHub - hitcxy/break_autokey

修改一下脚本（因为我是python3的）运行得到flag

flag：

flag{do_you_like_classical_cipher}

crypto2_月自圆

题目：

# -*- coding:utf-8 -*-
#Author: Lazzarofrom itertools import *
from random import *
from string import *def encrypt(m, a, si):c=""for i in range(len(m)):c+=hex(((ord(m[i])) * a + ord(next(si))) % 128)[2:].zfill(2)return cif __name__ == "__main__":m = '****************************************************flag{*************}'assert(len(m)==71)a = randint(50,100)salt = ''.join(sample(ascii_uppercase, 4))si = cycle(salt.lower())c=encrypt(m, a, si)print(c)#3472184e657e50561c481f5c1c4e1938163e154431015e13062c1b073d4e3a444f4a5c5c7a071919167b034e1c29566647600c4e1c2956661b6c1f50622f0016317e563546202a

程序逻辑：

随机生成盐值和一个整数用于加密；

salt生成迭代对象，可以无限循环salt本身的字符；

cycle()函数的作用

用**cycle()**循环迭代,**cycle()**函数重复循环一组值,可用它循环数据集标识符对数据集进行分组

sample(序列a,n):

在从序列a中随机抽取n个元素，并将n个元素生以list形式返回。

所以实际上每次进行encrypto()函数加密盐值的只有一个字符

而a的值很小，可以一起爆破

明文中给出了部分“flag”；在密文中可以对照的找到“flag”对应的密文是1c295666”

可以对照这段字符进行爆破：看加密之后是否于密文相同；注意是需要单个字符分别爆破；

因为每次加密取得是单个字符而且si在每几次加密中可能是不同字符的（假设salt = “aes”，那么第一次加密就是si=“a”，第二次加密si = “e”，以此类推）

爆破出各个字符满足的a，进行对比可以找到唯一满足情况的a

然后爆破salt，还是代入encrypto()函数进行把爆破，看密文是否对应，需要爆破"flag{"

因为不确定salt的大小；结果爆破出来：“JESQJ”

猜测应该是salt = “JESQ”

这样就可以求解flag了，还是运用爆破明文字符（可打印字符范围）；需要注意每次爆破出一个明文字符，si的值需要换到下一个，而next()函数的传参不能是非对象的，所以这里另外创建一个中间量sii，而在encrpto()函数中把next()函数去掉

代码实现：

from string import *
from random import *
from itertools import *c = "3472184e657e50561c481f5c1c4e1938163e154431015e13062c1b073d4e3a444f4a5c5c7a071919167b034e1c29566647600c4e1c2956661b6c1f50622f0016317e563546202a"
def encrypt(m, a, si):c=""for i in range(len(m)):c+=hex(((ord(m[i])) * a + ord(next(si))) % 128)[2:].zfill(2)return cdef encrypt2(m, a, si):c=""for i in range(len(m)):c+=hex(((ord(m[i])) * a + ord(si)) % 128)[2:].zfill(2)return c
#for语句中的if语句需要分别使用，同时运行的话因为si的值固定所以无法正确爆破
# for a in range(50,100):
#     for j in ascii_uppercase:
#         si = cycle(j.lower())# if encrypt("f",a,si) == "1c":#     print("f:",end="")#     print(a,j)# if encrypt("l",a,si) == "29":#     print("l:",end="")#     print(a,j)# if encrypt("a",a,si) == "56":#     print("a:",end="")#     print(a,j)
#"fla"爆破对比a = 67
# for j in ascii_uppercase:
#     si = cycle(j.lower())# if encrypt("f",67,si) == "1c":#     print(j)# if encrypt("l",67,si) == "29":#     print(j)    # if encrypt("a",67,si) == "56":#     print(j)# if encrypt("g",67,si) == "66":#     print(j)
#"flag"爆破salt得到JESQ
salt = "JESQ"
a = 67
si = cycle(salt.lower())
for i in range(0,len(c),2):temp = c[i:i+2]sii = next(si)for j in range(32,127):temp2 = encrypt2(chr(j),a,sii)if temp == temp2:print(chr(j),end="")
# now_is_7fad9fcb-d361-4964-821c-177c906b8d20_flag_is_flag{md5(now-salt)}
import hashlib
flag = "7fad9fcb-d361-4964-821c-177c906b8d20-JESQ" #这里太🐕了，后面加上"-JESQ"是真的有点考理解
print("")
print(hashlib.md5(flag.encode("utf8")).hexdigest())

crypto3_多少离怀

题目：

# -*- coding:utf-8 -*-
#Author: Lazzarofrom Crypto.Util.number import getPrime,isPrime
from math import gamma
import randomdef nextPrime(n):n += 2 if n & 1 else 1while not isPrime(n):n += 2return ndef getNewPrime():A = getPrime(512)B = nextPrime(A - random.randint(1e4,1e5))return nextPrime(gamma(B+2)%A)p = getNewPrime()
q = getNewPrime()
r = getNewPrime()
n = p * q ** 2 * r ** 3
e = 0x10001
c = pow(flag,e,n)#pA=6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723982789
#pB=6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723922147
#qA=7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477881291
#qB=7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477807457
#n=4451906216583258787166698210560165433649728830889954633721198623488802305844782492171757604711145165920462286487680020347239300947225371917344589502941576734875830871998499135120227347066586066943289430156378296665669974728569678779668142712266780949126509440672273927433367293606776081254094682033167575930701870261219046464773708974194213798032346187463443317770758989273370488582862531630356263732232300508706676725203199729764016766683870925164232508407363688370458877688991733322055785233669885166225464068579486683574954699370175267031949720967812902215635630884502987094547523361027411501285252862476410213277925430392164226297316310465146003494714018456407793759170649913180823814850170639706664167149612984905056804131124522209409607977589884632999710708045656852149371030046919242039957767777840304466948549383597755811307383659188216421501912831203089570725388153416013596114462069777713822433178099904907224119
#c=1996198968748552041728429851810599627895157160099076033250854211280074825148767841655949210593646824507865483166496070951130337321360509148527292165245205219296211294789087358959553387392928560150390604911087085313000622842025416521494799132969818997182731021267942464323979261593380113740152841984062184326431879167516288834455296913822921806893572566867611541664848820247889274979245086440402996661226884320574824077910315143756471444347386795428338020162169391827182914043434253974549636668126789355991920452920806351939782281969098470635517019120996509180703896707990501216102290302162705699788457579330150149320348175742131887213742989509004374645723471497302400169849858253644606818874098604333865973357374444445825761600866472906771935670261641342221394488068630591190697667016958881530367047928341661857241378511420562236766886349565409774340321441504290366223243635878057759623855735794209219474650425139791831374

hint：

注意伽马函数Γ(x)和阶乘x!的关系式

威尔逊定理

逻辑很简单：

随机生成素数pA，将pA减去一个随机在1e4到1e5之间的整数的下一个素数得到pB

素因子p = gamma(pB+2)%pA；同理q

而题目已给出pA，pB，qA，qB

错误示范：

原始想法：按着加密的算法再算一遍就好了啊，直接可以求得p和q

但是运行过程得到报错：math range error

搜一搜gamma函数；再结合hint和大佬wp；

$g a m m a (n) = (n - 1)!$

威尔逊定理：

$(p - 1)! = - 1 (m o d p)$

$(p - 2)! = 1 (m o d p)$

而题目中最后pq是等于 $gamma(pB+2)%pAgamma(pB+2)~\%~pA$

也就相当于 $(pB+1)!%pA(pB+1)!~\%~pA$

既然有求余符号就尽量往同余1的方向带

构造这样的同余式：

$(pA−2)!/(pB+2)!∗(pB+2)!≡1(modpA)(pA-2)!/(pB+2)!*(pB+2)!\equiv1~(mod~pA)$

设 $x = (p A - 2)! / (p B + 2)!$ ； $y = (p B + 2)!$

要求的y的大小；也就是求x与模数pA的逆元

而 $p A - 2$ 与 $p B + 2$ 相差 $1 e 4$ 到 $1 e 5$ 之间

他们两者相除得到：

$p B * (p A - 2 - p B) * . . . * (p A - 2)$

使用封装函数gmpy2.invert()即可求得逆元

注意这里遍历的时候需要每一个遍历量先进行求逆元；因为这样的阶乘还是太大了，中间如果有求余过程的话会简化很多计算过程

得到逆元也就是得到p和q；之后算r；使用 $n//p//q^2$

算fai_n的时候需要注意素因子的次方情况下，欧拉函数等于 $p^n-p^{n-1})$

代码实现：

import gmpy2
from Crypto.Util.number import *
from sympy.ntheory.generate import nextprimepA = 6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723982789
pB = 6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723922147
qA = 7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477881291
qB = 7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477807457
n = 4451906216583258787166698210560165433649728830889954633721198623488802305844782492171757604711145165920462286487680020347239300947225371917344589502941576734875830871998499135120227347066586066943289430156378296665669974728569678779668142712266780949126509440672273927433367293606776081254094682033167575930701870261219046464773708974194213798032346187463443317770758989273370488582862531630356263732232300508706676725203199729764016766683870925164232508407363688370458877688991733322055785233669885166225464068579486683574954699370175267031949720967812902215635630884502987094547523361027411501285252862476410213277925430392164226297316310465146003494714018456407793759170649913180823814850170639706664167149612984905056804131124522209409607977589884632999710708045656852149371030046919242039957767777840304466948549383597755811307383659188216421501912831203089570725388153416013596114462069777713822433178099904907224119
c = 1996198968748552041728429851810599627895157160099076033250854211280074825148767841655949210593646824507865483166496070951130337321360509148527292165245205219296211294789087358959553387392928560150390604911087085313000622842025416521494799132969818997182731021267942464323979261593380113740152841984062184326431879167516288834455296913822921806893572566867611541664848820247889274979245086440402996661226884320574824077910315143756471444347386795428338020162169391827182914043434253974549636668126789355991920452920806351939782281969098470635517019120996509180703896707990501216102290302162705699788457579330150149320348175742131887213742989509004374645723471497302400169849858253644606818874098604333865973357374444445825761600866472906771935670261641342221394488068630591190697667016958881530367047928341661857241378511420562236766886349565409774340321441504290366223243635878057759623855735794209219474650425139791831374
e = 65537
#注释部分是错误示范#由于math range error；得到的数字超过了数学范围，不能这样直接求解
# p = sympy.nextprime(gamma(pB+2)%pA)
# print(p)#由于pA，pB太大了，无法求阶乘
# jiec1 = 1
# jiec2 = 1
# for i in range(1,pA - 2):
#     jiec1 *= i
# for i in range(1,pB + 1):
#     jiec2 *= i 
# x = jiec1 // jiec2
# y = gmpy2.invert(x,pA)def getrealprime(a,b):difference = a - 2 - (b + 1)temp = 1for i in range(0,difference):temp = temp * gmpy2.invert(a-2-i,a) % areturn nextprime(temp)
p = getrealprime(pA,pB)
print(p)
q = getrealprime(qA,qB)
print(q)
r = gmpy2.iroot(n // p // q**2,3)[0]
fai_n = (p-1)*(q-1)*q*(r-1)*r**2
d = gmpy2.invert(e,fai_n)
m = pow(c,d,n)
print(long_to_bytes(m))

这是解题过程中借鉴的大佬wp：https://blog.csdn.net/hiahiachang/article/details/108800210
https://blog.csdn.net/Kr0ne/article/details/108833373?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522163503761916780366522347%2522%252C%2522scm%2522%253A%252220140713.130102334.pc%255Fall.%2522%257D&request_id=163503761916780366522347&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2_allfirst_rank_ecpm_v1~rank_v31_ecpm-3-108833373.first_rank_v2_pc_rank_v29&utm_term=ctfshow+%E6%9C%88%E9%A5%BC%E6%9D%AF2020&spm=1018.2226.3001.4187

查看全文

http://www.lbrq.cn/news/1453105.html