变异凯撒

可以看出其偏移量是对每个字符：从第一个字符的偏移量为5，第二个字符的偏移量为6……第n个字符的偏移量为4+n。偏移量依次递增。

ciphertext = 'afZ_r9VYfScOeO_UL^RWUc'
j = 5
for i in ciphertext:
    print(chr(ord(i) + j), end='')
    j += 1

Quoted-printable

*http://web.chacuo.net/charsetquotedprintable*

=E9=82=A3=E4=BD=A0=E4=B9=9F=E5=BE=88=E6=A3=92=E5=93=A6

Rabbit

RSA

公钥：(ne) 私钥：pqd

在一次RSA密钥对生成中，假设p=473398607161，q=4511491，e=17
求解出d作为flga提交

知识点：pqe求d

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/5/27 13:13
# @Author  : upload
# @File    : rsa求d.py
# @Software: PyCharm

import gmpy2
p =gmpy2.mpz(473398607161)
q =gmpy2.mpz(4511491)
e =gmpy2.mpz(17)
phi_n= (p - 1) * (q - 1)
    d = gmpy2.invert(e, phi_n)
print("d is:")
print (d)

尝试在线网站分解N

http://factordb.com/index.php?query=98554799767

rsarsa

知识点：pqe 求m

Math is cool! Use the RSA algorithm to decode the secret message, c, p, q, and e are parameters for the RSA algorithm.


p =  9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q =  11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
e =  65537
c =  83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034

Use RSA to find the secret message
import gmpy2
p = 9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q = 11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
e = 65537
c = 83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034
n = p * q
fn = (p - 1) * (q - 1)
d = gmpy2.invert(e, fn)
print(d)
m = pow(c,d,n)
print(m)

RSA1

知识点：**dp,dp泄露**

参数是为了让解密的时候更快速产生

import gmpy2
import binascii
def decrypt(dp,dq,p,q,c):
    InvQ = gmpy2.invert(q,p)
    mp = pow(c,dp,p)
    mq = pow(c,dq,q)
    m=(((mp-mq)*InvQ)%p)*q+mq
    print(m)
    print (binascii.unhexlify(hex(m)[2:]))
p = 8637633767257008567099653486541091171320491509433615447539162437911244175885667806398411790524083553445158113502227745206205327690939504032994699902053229
q = 12640674973996472769176047937170883420927050821480010581593137135372473880595613737337630629752577346147039284030082593490776630572584959954205336880228469
dp = 6500795702216834621109042351193261530650043841056252930930949663358625016881832840728066026150264693076109354874099841380454881716097778307268116910582929
dq = 783472263673553449019532580386470672380574033551303889137911760438881683674556098098256795673512201963002175438762767516968043599582527539160811120550041
c = 24722305403887382073567316467649080662631552905960229399079107995602154418176056335800638887527614164073530437657085079676157350205351945222989351316076486573599576041978339872265925062764318536089007310270278526159678937431903862892400747915525118983959970607934142974736675784325993445942031372107342103852
decrypt(dp,dq,p,q,c)

RSA3

知识点：共模攻击

image-20210527171832561

image-20210527172307452

import libnum 
from gmpy2 import invert
# 欧几里得算法
def egcd(a, b):
  if a == 0:
    return (b, 0, 1)
  else:
    g, y, x = egcd(b % a, a)
    return (g, x - (b // a) * y, y)

def main():
  n = 22708078815885011462462049064339185898712439277226831073457888403129378547350292420267016551819052430779004755846649044001024141485283286483130702616057274698473611149508798869706347501931583117632710700787228016480127677393649929530416598686027354216422565934459015161927613607902831542857977859612596282353679327773303727004407262197231586324599181983572622404590354084541788062262164510140605868122410388090174420147752408554129789760902300898046273909007852818474030770699647647363015102118956737673941354217692696044969695308506436573142565573487583507037356944848039864382339216266670673567488871508925311154801
  c1 = 22322035275663237041646893770451933509324701913484303338076210603542612758956262869640822486470121149424485571361007421293675516338822195280313794991136048140918842471219840263536338886250492682739436410013436651161720725855484866690084788721349555662019879081501113222996123305533009325964377798892703161521852805956811219563883312896330156298621674684353919547558127920925706842808914762199011054955816534977675267395009575347820387073483928425066536361482774892370969520740304287456555508933372782327506569010772537497541764311429052216291198932092617792645253901478910801592878203564861118912045464959832566051361
  c2 = 18702010045187015556548691642394982835669262147230212731309938675226458555210425972429418449273410535387985931036711854265623905066805665751803269106880746769003478900791099590239513925449748814075904017471585572848473556490565450062664706449128415834787961947266259789785962922238701134079720414228414066193071495304612341052987455615930023536823801499269773357186087452747500840640419365011554421183037505653461286732740983702740822671148045619497667184586123657285604061875653909567822328914065337797733444640351518775487649819978262363617265797982843179630888729407238496650987720428708217115257989007867331698397
  e1 = 11187289
  e2 = 9647291
  s = egcd(e1, e2)
  s1 = s[1]
  s2 = s[2]
  # 求模反元素
  if s1<0:
    s1 = - s1
    c1 = invert(c1, n)
  elif s2<0:
    s2 = - s2
    c2 = invert(c2, n)

  m = pow(c1,s1,n)*pow(c2,s2,n) % n
  print (libnum.n2s(m))

if __name__ == '__main__':
  main()

RSA2

知识点：dp泄露

在这里插入图片描述

遍历x求p！

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/5/27 17:44
# @Author  : upload
# @File    : mythink.py
# @Software: PyCharm
import binascii


import gmpy2 as gp

e = 65537
n = 248254007851526241177721526698901802985832766176221609612258877371620580060433101538328030305219918697643619814200930679612109885533801335348445023751670478437073055544724280684733298051599167660303645183146161497485358633681492129668802402065797789905550489547645118787266601929429724133167768465309665906113
dp = 905074498052346904643025132879518330691925174573054004621877253318682675055421970943552016695528560364834446303196939207056642927148093290374440210503657

c = 140423670976252696807533673586209400575664282100684119784203527124521188996403826597436883766041879067494280957410201958935737360380801845453829293997433414188838725751796261702622028587211560353362847191060306578510511380965162133472698713063592621028959167072781482562673683090590521214218071160287665180751


for i in range(1,e):                   #在范围(1,e)之间进行遍历
    if(dp*e-1)%i == 0:
        if n%(((dp*e-1)//i)+1) == 0:   #存在p，使得n能被p整除
            p=((dp*e-1)//i)+1
            q=n//(((dp*e-1)//i)+1)
            phi=(q-1)*(p-1)            #欧拉定理
            d=gp.invert(e,phi)         #求模逆
            m=pow(c,d,n)               #快速求幂取模运算

print(m)                               #10进制明文
print('------------')
print(hex(m)[2:])                      #16进制明文
print('------------')
print (binascii.unhexlify(hex(m)[2:]))     #16进制转文本

flag{37995588256861228614165223347687}

我符了 感觉不用dp攻击都可以！ 有n直接由 pq 又有e，就有d 有d就完了啊！

import binascii

import gmpy2
p = 13468634736343473907717969603434376212206335187555458742257940406618189481177835992217885676243155145465521141546915941147336786447889325606555333350540003
q = 18432009829596386103558375461387837845170621179295293289126504231317130550979989727125205467379713835047300158256398009229511746203459540859429194971855371
e = 65537
c = 140423670976252696807533673586209400575664282100684119784203527124521188996403826597436883766041879067494280957410201958935737360380801845453829293997433414188838725751796261702622028587211560353362847191060306578510511380965162133472698713063592621028959167072781482562673683090590521214218071160287665180751
n = p * q
print(n)
fn = (p - 1) * (q - 1)
d = gmpy2.invert(e, fn)
print(d)
m = pow(c,d,n)
print(m)
print(hex(m)[2:])
print('------------')
print (binascii.unhexlify(hex(m)[2:]))     #16进制转文本

异性相吸

知识点:xor

#!/usr/bin/env python2
# -*- coding: utf-8 -*-
# @Time    : 2021/5/27 18:08
# @Author  : upload
# @File    : xor.py
# @Software: PyCharm
a = '0110000101110011011000010110010001110011011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011100010111011101100101011100110111000101100110'
b = '0000011100011111000000000000001100001000000001000001001001010101000000110001000001010100010110000100101101011100010110000100101001010110010100110100010001010010000000110100010000000010010110000100011000000110010101000100011100000101010101100100011101010111010001000001001001011101010010100001010000011011'
c = ''

for i in range(len(a)):
    if(a[i] == b[i]):
        c+='0'
    else:
        c+='1'
print(c)

RSA

知识点：公钥加签 给公钥求m

cmd：

yafu-x64 factor(86934482296048119190666062003494800588905656017203025617216654058378322103517)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# @Time    : 2021/5/27 18:37
# @Author  : upload
# @File    : 带文件.py
# @Software: PyCharm
import rsa

e= 65537
n= 86934482296048119190666062003494800588905656017203025617216654058378322103517
p= 285960468890451637935629440372639283459
q= 304008741604601924494328155975272418463
d= 81176168860169991027846870170527607562179635470395365333547868786951080991441

key = rsa.PrivateKey(n,e,d,q,p)         #在pkcs标准中,pkcs#1规定,私钥包含(n,e,d,p,q)

with open("D:\oneDrive云存储\OneDrive\桌面\密码冲/flag.txt","rb") as f:  #以二进制读模式，读取密文
    f = f.read()
    print(rsa.decrypt(f,key))           # f:公钥加密结果  key:私钥

还原大师

知识点：我就叫他哈希碰撞了！

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/5/27 18:43
# @Author  : upload
# @File    : md5 换源.py
# @Software: PyCharm
import hashlib

#print hashlib.md5(s).hexdigest().upper()
k = 'TASC?O3RJMV?WDJKX?ZM'                    #要还原的明文
for i in range(26):
	temp1 = k.replace('?',str(chr(65+i)),1)
	for j in range(26):
		temp2 = temp1.replace('?',chr(65+j),1)
		for n in range(26):
			temp3 = temp2.replace('?',chr(65+n),1)
			s = hashlib.md5(temp3.encode('utf8')).hexdigest().upper()#注意大小写
			if s[:4] == 'E903':    #检查元素
				print (s)       #输出密文

RSAROLL

知识点：正常的rsa 给了 n e

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/5/27 19:03
# @Author  : upload
# @File    : rsarool.py
# @Software: PyCharm

import gmpy2
p = 49891
q = 18443
e = 19
n =  920139713
phi_n= (p - 1) * (q - 1)
d = gmpy2.invert(e, phi_n)
c = [704796792,752211152,274704164,18414022,368270835,483295235,263072905,459788476,483295235,459788476,663551792,475206804,459788476,428313374,475206804,459788476,425392137,704796792,458265677,341524652,483295235,534149509,425392137,428313374,425392137,341524652,458265677,263072905,483295235,828509797,341524652,425392137,475206804,428313374,483295235,475206804,459788476,306220148]

re = ''
for i in c:
    m = pow(i,d,n)
    re += str(chr(m))
print(re)