cry6

[WUSTCTF2020]B@se

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/6 17:27
# @Author  : upload
# @File    : 111111111111.py
# @Software: PyCharm
import binascii
str="JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs****kxyz012789+/"
ciper="MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbVD"#(==没有用)
import string
import binascii
for i in string.ascii_letters+string.digits:
    if i not in str:
        print(i)
import itertools
s=['j','u','3','4']
for i in itertools.permutations(s,4):
    ss="JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs"+"".join(i)+"kxyz012789+/"
    bins = ""
    for j in ciper:
      bins+=bin(ss.index(j))[2:].zfill(6)
    print(binascii.unhexlify(hex(eval("0b"+bins))[2:-1]))

RSA & what

知识点：RSA共模攻击 + Base64隐写

RSA共模攻击

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/6 17:27
# @Author  : upload
# @File    : 111111111111.py
# @Software: PyCharm
import binascii
from Crypto.Util.number import*
import base64

#共模攻击
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 CMA(n,e1,e2,c1,c2):
    s = egcd(e1, e2)
    s1 = s[1]
    s2 = s[2]
    if s1<0:
        s1 = - s1
        c1 = inverse(c1, n)
    elif s2<0:
        s2 = - s2
        c2 = inverse(c2, n)
    m = pow(c1,s1,n)*pow(c2,s2,n) % n
    return m

f1=open("D:/oneDrive云存储/OneDrive/桌面/cry/rsawhat/HUB1")
f2=open("D:/oneDrive云存储/OneDrive/桌面/cry/rsawhat/HUB2")
N=f1.readline()
N=f2.readline()
e1,e2=f1.readline(),f2.readline()
# print(e1,e2)
f1.readline()
f2.readline()
c1,c2=f1.readline(),f2.readline()
ans=b''
cnt=0
while len(c1)!=0:
    cnt+=1
    ans+=long_to_bytes(CMA(int(N),int(e1),int(e2),int(c1),int(c2)))
    #print(base64.b64decode(temp))
    c1,c2=f1.readline(),f2.readline()
temp=b''
M=b''
print(ans)
for i in ans:
    k=long_to_bytes(i)
    #print(i," ",end="")
    if k==b'\n':
        M+=base64.b64decode(temp)
        temp=b''
        continue
    temp+=k
print(M)

Base64隐写

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/5/31 17:11
# @Author  : upload
# @File    : test.py
# @Software: PyCharm
from Crypto.Util.number import*
import base64
c = b'VEhJUz==\nRkxBR3==\nSVN=\nSElEREVOLo==\nQ0FO\nWU9V\nRklORM==\nSVT=\nT1VUP4==\nRE8=\nWU9V\nS05PV9==\nQkFTRTY0P5==\nWW91bmdD\nVEhJTku=\nWU9V\nQVJF\nTk9U\nVEhBVE==\nRkFNSUxJQVI=\nV0lUSO==\nQkFTRTY0Lh==\nQmFzZTY0\naXO=\nYW==\nZ3JvdXA=\nb2b=\nc2ltaWxhcn==\nYmluYXJ5LXRvLXRleHR=\nZW5jb2Rpbme=\nc2NoZW1lc0==\ndGhhdD==\ncmVwcmVzZW50\nYmluYXJ5\nZGF0YW==\naW5=\nYW6=\nQVNDSUl=\nc3RyaW5n\nZm9ybWF0\nYnk=\ndHJhbnNsYXRpbmd=\naXS=\naW50b1==\nYT==\ncmFkaXgtNjQ=\ncmVwcmVzZW50YXRpb24u\nVGhl\ndGVybc==\nQmFzZTY0\nb3JpZ2luYXRlc8==\nZnJvbd==\nYY==\nc3BlY2lmaWN=\nTUlNRT==\nY29udGVudI==\ndHJhbnNmZXI=\nZW5jb2Rpbmcu\nVGhl\ncGFydGljdWxhct==\nc2V0\nb2b=\nNjR=\nY2hhcmFjdGVyc5==\nY2hvc2Vu\ndG+=\ncmVwcmVzZW50\ndGhl\nNjQ=\ncGxhY2UtdmFsdWVz\nZm9y\ndGhl\nYmFzZd==\ndmFyaWVz\nYmV0d2Vlbt==\naW1wbGVtZW50YXRpb25zLp==\nVGhl\nZ2VuZXJhbI==\nc3RyYXRlZ3n=\naXO=\ndG9=\nY2hvb3Nl\nNjR=\nY2hhcmFjdGVyc5==\ndGhhdA==\nYXJl\nYm90aN==\nbWVtYmVyc5==\nb2a=\nYS==\nc3Vic2V0\nY29tbW9u\ndG8=\nbW9zdM==\nZW5jb2RpbmdzLA==\nYW5k\nYWxzb8==\ncHJpbnRhYmxlLg==\nVGhpc9==\nY29tYmluYXRpb25=\nbGVhdmVz\ndGhl\nZGF0YW==\ndW5saWtlbHk=\ndG/=\nYmV=\nbW9kaWZpZWS=\naW5=\ndHJhbnNpdE==\ndGhyb3VnaN==\naW5mb3JtYXRpb26=\nc3lzdGVtcyw=\nc3VjaN==\nYXM=\nRS1tYWlsLD==\ndGhhdA==\nd2VyZQ==\ndHJhZGl0aW9uYWxseQ==\nbm90\nOC1iaXQ=\nY2xlYW4uWzFd\nRm9y\nZXhhbXBsZSw=\nTUlNRSdz\nQmFzZTY0\naW1wbGVtZW50YXRpb24=\ndXNlcw==\nQahDWiw=\nYahDeiw=\nYW5k\nMKhDOQ==\nZm9y\ndGhl\nZmlyc3Q=\nNjI=\ndmFsdWVzLg==\nT3RoZXI=\ndmFyaWF0aW9ucw==\nc2hhcmU=\ndGhpcw==\ncHJvcGVydHk=\nYnV0\nZGlmZmVy\naW4=\ndGhl\nc3ltYm9scw==\nY2hvc2Vu\nZm9y\ndGhl\nbGFzdA==\ndHdv\ndmFsdWVzOw==\nYW4=\nZXhhbXBsZQ==\naXM=\nVVRGLTcu'

def get_base64_diff_value(s1, s2):
    base64chars = b'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
    res = 0
    for i in range(len(s2)):
        if s1[i] != s2[i]:
            return abs(base64chars.index(s1[i]) - base64chars.index(s2[i]))
    return res

def solve_stego():
    line=b''
    bin_str=''
    for i in c:
        k=long_to_bytes(i)
        if k==b'\n':
            steg_line = line
            norm_line = base64.b64encode(base64.b64decode(line))
            diff = get_base64_diff_value(steg_line, norm_line)
            #print(diff)
            pads_num = steg_line.count(b'=')
            if diff:
                bin_str += bin(diff)[2:].zfill(pads_num * 2)
            else:
                bin_str += '0' * pads_num * 2
            print(goflag(bin_str))
            line=b''
            continue
        line+=k


def goflag(bin_str):
    res_str = ''
    for i in range(0, len(bin_str), 8):
        res_str += chr(int(bin_str[i:i + 8], 2))
    print('flag{' + res_str + '}')
    return res_str


if __name__ == '__main__':
    solve_stego()

[WUSTCTF2020]大数计算

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/6 17:27
# @Author  : upload
# @File    : 111111111111.py
# @Software: PyCharm
import binascii

sum = 1
for i in range(1,2021):
    sum = sum*i


a =   hex(int(str(sum)[0:8]))[2:]
print(a)
# print(str(sum)[0:8])

b =   hex(int(str(520**1314+2333**666)[0:8]))[2:]
print(b)
# print(str(520**1314+2333**666)[0:8])

c =  hex(int(str(80538738812075974+80435758145817515+12602123297335631)[0:8]))[2:]
print(c)
# print(str(80538738812075974+80435758145817515+12602123297335631)[0:8])

d = hex((22*22+36)*1314)[2:]
print(d)
# print((22*22+36)*1314)
print('flag{'+str(a)+'-'+str(b)+'-'+str(c)+'-'+str(d)+'}')

[HDCTF2019]together

知识点：共模攻击 base64解码

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/6 17:27
# @Author  : upload
# @File    : 111111111111.py
# @Software: PyCharm
import binascii

import base64
from Crypto.PublicKey import RSA
from Crypto.Util.number import inverse, bytes_to_long, long_to_bytes

# path = r'D:\oneDrive云存储\OneDrive\桌面\cry\attachment (1)\pubkey1.pem'
#
# for i in range(2):
#
#     with open(path) as f:
#
#         key = RSA.import_key(f.read())
#
#         # print('e=%d' % key.e)
#         #
#         # print('n=%d' % key.n)
#     f.close()
#     path = r'D:\oneDrive云存储\OneDrive\桌面\cry\attachment (1)\pubkey2.pem'
# f.close()
#
# filename = 'D:/oneDrive云存储/OneDrive/桌面/cry/attachment (1)/myflag1'
# for i in range(2):
#     with open(filename,'r') as f:
#         c = f.readline()
#         # print(c)
#     print()
#     filename = 'D:/oneDrive云存储/OneDrive/桌面/cry/attachment (1)/myflag2'
#
c1 = 'R3Noy6r3WLItytAmb4FmHEygoilucEEZbO9ZYXx5JN03HNpBLDx7fXd2fl+UL5+11RCs/y0qlTGURWWDtG66eNLzGwNpAKiVj6I7RtUJl2Pcm3NvFeAFwI9UsVREyh7zIV6sI9ZP8l/2GVDorLAz5ULW+f0OINGhJmZm8FL/aDnlfTElhQ87LPicWpXYoMtyr6WrxjK6Ontn8BqCt0EjQ7TeXZhxIH9VTPWjDmFdmOqaqdVIT+LZemTgLNESwM5nn4g5S3aFDFwj1YiDYl0/+8etvKfOrfoKOwR0CxsRHagwdUUTES8EcHLmMGCxCkDZn3SzmmA6Nb3lgLeSgG8P1A=='
c2 = 'O+rRCXI3aTB6P1rYIOPUdalUp6ujpwEq4I20CoWA+HIL8xxGtqY6N5gpr0guZv9ZgOEAMFnBxOqMdVNnB9GgnhmXtt1ZWydPqIcHvlfwpd/Lyd0XSjXnjaz3P3vOQvR71cD/uXyBA0XPzmnTIMgEhuGJVFm8min0L/2qI7wg/Z7w1+4mOmi655JIXeCiG23ukDv6l9bZuqfGvWCa1KKXWDP31nLbp0ZN2obUs6jEAa1qVTaX6M4My+sks+0VvHATrAUuCrmMwVEivqIJ/nS6ymGVERN6Ohnzyr168knEBKOVj0FAOx3YLfppMM+XbOGHeqdKJRLpMvqFXDMGQInT3w=='

c1 = '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'
c2 = '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'

c1 = int(c1,16)
c2 = int(c2,16)
e1 = 2333
e2 = 23333

n = 14853081277902411240991719582265437298941606850989432655928075747449227799832389574251190347654658701773951599098366248661597113015221566041305501996451638624389417055956926238595947885740084994809382932733556986107653499144588614105694518150594105711438983069306254763078820574239989253573144558449346681620784979079971559976102366527270867527423001083169127402157598183442923364480383742653117285643026319914244072975557200353546060352744263637867557162046429886176035616570590229646013789737629785488326501654202429466891022723268768841320111152381619260637023031430545168618446134188815113100443559425057634959299

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 CMA(n,e1,e2,c1,c2):
    s = egcd(e1, e2)
    s1 = s[1]
    s2 = s[2]
    if s1<0:
        s1 = - s1
        c1 = inverse(c1, n)
    elif s2<0:
        s2 = - s2
        c2 = inverse(c2, n)
    m = pow(c1,s1,n)*pow(c2,s2,n) % n
    # print(1)
    print(binascii.unhexlify(hex(m)[2:]))

    return m

if __name__ == '__main__':
    # print(c1,c2)
    CMA(n,e1,e2,c1,c2)

from gmpy2 import iroot,invert

n = 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

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)



c1 = 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

c2 = 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

e1 = 2333

e2 = 23333

s = egcd(e1,e2)

s1 = s[1]

s2 = s[2]

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 hex(m)[2:].decode('hex')

[MRCTF2020]babyRSA

知识点：修改的RSA

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/6 17:27
# @Author  : upload
# @File    : 111111111111.py
import gmpy2
import libnum
import sympy
from Crypto.Util.number import long_to_bytes

P_p = 206027926847308612719677572554991143421
P_factor = 213671742765908980787116579976289600595864704574134469173111790965233629909513884704158446946409910475727584342641848597858942209151114627306286393390259700239698869487469080881267182803062488043469138252786381822646126962323295676431679988602406971858136496624861228526070581338082202663895710929460596143281673761666804565161435963957655012011051936180536581488499059517946308650135300428672486819645279969693519039407892941672784362868653243632727928279698588177694171797254644864554162848696210763681197279758130811723700154618280764123396312330032986093579531909363210692564988076206283296967165522152288770019720928264542910922693728918198338839
Q_1 =  103766439849465588084625049495793857634556517064563488433148224524638105971161051763127718438062862548184814747601299494052813662851459740127499557785398714481909461631996020048315790167967699932967974484481209879664173009585231469785141628982021847883945871201430155071257803163523612863113967495969578605521
Q_2 =  151010734276916939790591461278981486442548035032350797306496105136358723586953123484087860176438629843688462671681777513652947555325607414858514566053513243083627810686084890261120641161987614435114887565491866120507844566210561620503961205851409386041194326728437073995372322433035153519757017396063066469743
sub_Q =  168992529793593315757895995101430241994953638330919314800130536809801824971112039572562389449584350643924391984800978193707795909956472992631004290479273525116959461856227262232600089176950810729475058260332177626961286009876630340945093629959302803189668904123890991069113826241497783666995751391361028949651
Ciphertext =  1709187240516367141460862187749451047644094885791761673574674330840842792189795049968394122216854491757922647656430908587059997070488674220330847871811836724541907666983042376216411561826640060734307013458794925025684062804589439843027290282034999617915124231838524593607080377300985152179828199569474241678651559771763395596697140206072537688129790126472053987391538280007082203006348029125729650207661362371936196789562658458778312533505938858959644541233578654340925901963957980047639114170033936570060250438906130591377904182111622236567507022711176457301476543461600524993045300728432815672077399879668276471832



def p_prime(p):
    while 1:
        p=p-2
        if(sympy.isprime(p)):
            return p



Q = pow(sub_Q,Q_2,Q_1)
_Q = sympy.nextprime(Q)
print(12333)

if __name__ == '__main__':

    n =1
    e = 65537
    P = [0 for i in range(17)]
    P[9] = P_p
    for i in range(10,17):
        P[i]=sympy.nextprime(P[i-1])
    for i in range(9,0,-1):
        P[i-1]=p_prime(P[i])
    for i in range(17):
        n*=P[i]

    fn =1
    for i in range(17):
        fn = fn*(P[i]-1)
    print(fn)
    d =  gmpy2.invert(e, fn)
    _P1 = pow(P_factor,d,n)
    _P = sympy.nextprime(_P1)

    fn2 = (_P-1)*(_Q-1)
    d2 = gmpy2.invert(e, fn2)
    _M = pow(Ciphertext,d2,_P*_Q)
    print(str(long_to_bytes(_M)).replace('MRCTF','flag'))

我们知道欧拉函数的定义：

ϕ(n1n2…nn)=(n1−1)(n2−1)…(nn−1)ϕ(n1n2…nn)=(n1−1)(n2−1)…(nn−1)

这样就可以解出问题中p的值，再根据nextprime函数就可以求出_P,有了_P和_Q直接进行解密就可以了，

因为 RSA解密的时候需要欧拉函数！来求d! n是由大素数构成！ 所以满足欧拉公式！那我们可以求出欧拉函数！就有了D!

[BJDCTF2020]Polybius(波利比奥斯方阵密码)

知识点：波利比奥斯方阵密码

import itertools
s="aeoiu"
sumresult=[]
numsumresult=[]
ciper="ouauuuoooeeaaiaeauieuooeeiea"
for i in itertools.permutations(s,5):#找出所有全排列
    sumresult.append("".join(i))
for i in sumresult:
    temp=""
    for j in ciper:
        temp+=str(i.index(j)+1)
    numsumresult.append(temp)
for i in numsumresult:
    ans_=""
    for j in range(0, len(i),2):
        xx=(int(i[j])-1)*5+int(i[j+1])+96
        if xx>ord('i'):
            xx+=1
        ans_+=chr(xx)
    print(ans_)

[AFCTF2018]BASE

大base

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/1 15:16
# @Author  : upload
# @File    : [BJDCTF2020]easyrsa.py
# @Software: PyCharm
from base64 import *
s = ""
with open('D:\\oneDrive云存储\\OneDrive\\桌面\\misc/flag_encode.txt', 'r') as file:
    s = bytes(file.read(),'ascii')
    file.close()

lis1 = [s]
lis2 = []
lis3 = []
lis4 = []
while(1):
	for a in lis1:
		ok = 0
		try:
			lis2.append(b64decode(a).decode('ascii'))
			ok = 1
		except:
			pass
		try:
			lis2.append(b32decode(a).decode('ascii'))
			ok = 1
		except:
			pass
		try:
			lis2.append(b16decode(a).decode('ascii'))
			ok = 1
		except:
			pass
		if not ok:
			lis3.append(a)
	if not len(lis2):
		break
	lis1=lis2.copy()
	lis2.clear()
for a in range(0,len(lis3)):
	ok = 1
	for b in lis3[a]:
		if ord(b)>126 or ord(b)<32:
			ok = 0
			break
	if ok:
		lis4.append(lis3[a])
print(lis4)

[NPUCTF2020]Classical Cipher

知识点：变异密码！

'''
from z3 import *

a = Int('a')
b = Int('b')
s = Solver()
s.add((10*a+b)%26==15)
s.add((4*a+b)%26==21)
s.add((24*a+b)%26==1)
if s.check()==sat:
    print s.model()
'''
import gmpy2

a = 25
b = 25
n = gmpy2.invert(a,26)
c = 'gsv_pvb_rh_zgyzhs'
m = ''
l = 'abcdefghijklmnopqrstuvwxyz'
for i in c:
    if i=='_':
        m += i
    else:
        m += l[((ord(i)-97-b)*n)%26]
print(m)

EasyProgram

知识点：代码

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/1 15:16
# @Author  : upload
# @File    : [BJDCTF2020]easyrsa.py
# @Software: PyCharm
def filehex(file):
    fhex=[]
    f = open(file,'rb')
    ff = f.read().hex()

    for i in range(0,len(str(ff)),2):
        fhex.append(int(str(ff)[i:i+2],16))
    return fhex

flagx2=filehex('D:/oneDrive云存储/OneDrive/桌面/misc/附件/file.txt')
print(flagx2)

#使用010edit 读取'file.txt'文件16进制
flagx=[0x00,0xBA,0x8F,0x11,0x2B,0x22,0x9F,0x51,0xA1,0x2F,0xAB,0xB7,0x4B,0xD7,0x3F,0xEF,0xE1,0xB5,0x13,0xBE,0xC4,0xD4,0x5D,0x03,0xD9,0x00,0x7A,0xCA,0x1D,0x51,0xA4,0x73,0xB5,0xEF,0x3D,0x9B,0x31,0xB3]

s=[]
t=[]
key='whoami'
j=0
for i in range(0,256):
    s.append(i)

for i in range(0,256):
    t.append(key[i % len(key)])

for i in range(0,256):
     j=(j+int(s[i])+int(ord(t[i])))%256
     s[i], s[j] = s[j], s[i]

i=0;j=0;x=0
for m in range(0,38):
    i=(i + 1)%(256)
    j=(j + s[i])%(256)
    s[i],s[j] = s[j],s[i]
    x=(s[i] + (s[j]%(256)))%(256)
    flagx[m] = flagx[m] ^ s[x]
    flagx2[m]=flagx2[m]^s[x]

print(''.join(chr(flagx[i]) for i in range(0,38)))
print(''.join(chr(flagx2[i]) for i in range(0,38)))

[NPUCTF2020]EzRSA

知识点：e不是素数

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/1 15:16
# @Author  : upload
# @File    : [BJDCTF2020]easyrsa.py
# @Software: PyCharm

import gmpy2
from Crypto.Util.number import long_to_bytes

q = 106021448991021391444550749375115277080844281746248845802565680557785009341952320484175568763707424932172033597514861602114171459176440279045761846695231788376075050452154924141266290931413542110639081792550648106240966552406813059396358355737185354885474455248579946190266152416149137616855791805617206153497
p = 161136651053130509602530659420755324119806487925813087617466818245407407797561810253722204813002837916779909309520498985459703212021249251124954613236122142746302911323565396331355397916764254680629384957057354297855676493062493901977415968666512459829211010720514167083018352796496733697235524845188512914793

c = 3738960639194737957667684143565005503596276451617922474669745529299929395507971435311181578387223323429323286927370576955078618335757508161263585164126047545413028829873269342924092339298957635079736446851837414357757312525158356579607212496060244403765822636515347192211817658170822313646743520831977673861869637519843133863288550058359429455052676323196728280408508614527953057214779165450356577820378810467527006377296194102671360302059901897977339728292345132827184227155061326328585640019916328847372295754472832318258636054663091475801235050657401857262960415898483713074139212596685365780269667500271108538319


fn = (p-1)*(q-1)
e = 54722
n = p*q
# print(n)
d = gmpy2.invert(e // 2, fn)
m_2 = pow(c, int(d), n)

# e除以2后 求的是m的平方！所以要对m再开次方
flag = long_to_bytes(gmpy2.isqrt(m_2))
print(str(flag).replace('NPUCTF','flag'))

[De1CTF2019]babyrsa

知识点：好题 rsa多知识点嵌套

最后两个方程分解flag没看懂！也真不想看了！‘

1中国剩余定理！ 看见就给了 N和c

2 小e攻击 ！

3 直接分解 n

4 两个方程求flag

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/1 14:47
# @Author  : upload
# @File    : RSA4.py
# @Software: PyCharm

import gmpy2
import math
def  merge(a1,n1,a2,n2):
    d = math.gcd(n1,n2)
    c = a2-a1
    if c%d!=0:
        return 0
    c = (c%n2+n2)%n2
    c = c//d
    n1 = n1//d
    n2 = n2//d
    c *= gmpy2.invert(n1,n2)
    c %= n2
    c *= n1*d
    c += a1
    global n3
    global a3
    n3 = n1*n2*d
    a3 = (c%n3+n3)%n3
    return 1
def exCRT(a,n):
    a1=a[0]
    n1=n[0]
    le= len(a)
    for i in range(1,le):
        a2 = a[i]
        n2=n[i]
        if not merge(a1,n1,a2,n2):
            return -1
        a1 = a3
        n1 = n3
    global mod
    mod=n1
    return (a1%n1+n1)%n1
def exCRT_getequation(a,n):
    a1=a[0]
    n1=n[0]
    le= len(a)
    for i in range(1,le):
        a2 = a[i]
        n2=n[i]
        if not merge(a1,n1,a2,n2):
            return -1
        a1 = a3
        n1 = n3
    return (a1,n1)
#a为余数列表
#n为模数列表
n =  [20129615352491765499340112943188317180548761597861300847305827141510465619670536844634558246439230371658836928103063432870245707180355907194284861510906071265352409579441048101084995923962148527097370705452070577098780246282820065573711015664291991372085157016901209114191068574208680397710042842835940428451949500607613634682684113208766694028789275748528254287705759528498986306494267817198340658241873024800336013946294891687591013414935237821291805123285905335762719823771647853378892868896078424572232934360940672962436849523915563328779942134504499568866135266628078485232098208237036724121481835035731201383423, 31221650155627849964466413749414700613823841060149524451234901677160009099014018926581094879840097248543411980533066831976617023676225625067854003317018794041723612556008471579060428898117790587991055681380408263382761841625714415879087478072771968160384909919958010983669368360788505288855946124159513118847747998656422521414980295212646675850690937883764000571667574381419144372824211798018586804674824564606122592483286575800685232128273820087791811663878057827386379787882962763290066072231248814920468264741654086011072638211075445447843691049847262485759393290853117072868406861840793895816215956869523289231421, 29944537515397953361520922774124192605524711306753835303703478890414163510777460559798334313021216389356251874917792007638299225821018849648520673813786772452822809546571129816310207232883239771324122884804993418958309460009406342872173189008449237959577469114158991202433476710581356243815713762802478454390273808377430685157110095496727966308001254107517967559384019734279861840997239176254236069001453544559786063915970071130087811123912044312219535513880663913831358790376650439083660611831156205113873793106880255882114422025746986403355066996567909581710647746463994280444700922867397754748628425967488232530303, 25703437855600135215185778453583925446912731661604054184163883272265503323016295700357253105301146726667897497435532579974951478354570415554221401778536104737296154316056314039449116386494323668483749833147800557403368489542273169489080222009368903993658498263905567516798684211462607069796613434661148186901892016282065916190920443378756167250809872483501712225782004396969996983057423942607174314132598421269169722518224478248836881076484639837343079324636997145199835034833367743079935361276149990997875905313642775214486046381368619638551892292787783137622261433528915269333426768947358552919740901860982679180791]
c =  [19131432661217908470262338421299691998526157790583544156741981238822158563988520225986915234570037383888112724408392918113942721994125505014727545946133307329781747600302829588248042922635714391033431930411180545085316438084317927348705241927570432757892985091396044950085462429575440060652967253845041398399648442340042970814415571904057667028157512971079384601724816308078631844480110201787343583073815186771790477712040051157180318804422120472007636722063989315320863580631330647116993819777750684150950416298085261478841177681677867236865666207391847046483954029213495373613490690687473081930148461830425717614569, 15341898433226638235160072029875733826956799982958107910250055958334922460202554924743144122170018355117452459472017133614642242411479849369061482860570279863692425621526056862808425135267608544855833358314071200687340442512856575278712986641573012456729402660597339609443771145347181268285050728925993518704899005416187250003304581230701444705157412790787027926810710998646191467130550713600765898234392350153965811595060656753711278308005193370936296124790772689433773414703645703910742193898471800081321469055211709339846392500706523670145259024267858368216902176489814789679472227343363035428541915118378163012031, 18715065071648040017967211297231106538139985087685358555650567057715550586464814763683688299037897182845007578571401359061213777645114414642903077003568155508465819628553747173244235936586812445440095450755154357646737087071605811984163416590278352605433362327949048243722556262979909488202442530307505819371594747936223835233586945423522256938701002370646382097846105014981763307729234675737702252155130837154876831885888669150418885088089324534892506199724486783446267336789872782137895552509353583305880144947714110009893134162185382309992604435664777436197587312317224862723813510974493087450281755452428746194446, 2282284561224858293138480447463319262474918847630148770112472703128549032592187797289965592615199709857879008271766433462032328498580340968871260189669707518557157836592424973257334362931639831072584824103123486522582531666152363874396482744561758133655406410364442174983227005501860927820871260711861008830120617056883514525798709601744088135999465598338635794275123149165498933580159945032363880613524921913023341209439657145962332213468573402863796920571812418200814817086234262280338221161622789516829363805084715652121739036183264026120868756523770196284142271849879003202190966150390061195469351716819539183797]
p_4=exCRT(c,n)
p=gmpy2.iroot(p_4,4)[0]
#中国剩余定理
ee1 = 42
ee2 = 3
ce1 =  45722651786340123946960815003059322528810481841378247280642868553607692149509126962872583037142461398806689489141741494974836882341505234255325683219092163052843461632338442529011502378931140356111756932712822516814023166068902569458299933391973504078898958921809723346229893913662577294963528318424676803942288386430172430880307619748186863890050113934573820505570928109017842647598266634344447182347849367714564686341871007505886728393751147033556889217604647355628557502208364412269944908011305064122941446516990168924709684092200183860653173856272384
ce2 =  13908468332333567158469136439932325992349696889129103935400760239319454409539725389747059213835238373047899198211128689374049729578146875309231962936554403287882999967840346216695208424582739777034261079550395918048421086843927009452479936045850799096750074359160775182238980989229190157551197830879877097703347301072427149474991803868325769967332356950863518504965486565464059770451458557744949735282131727956056279292800694203866167270268988437389945703117070604488999247750139568614939965885211276821987586882908159585863514561191905040244967655444219603287214405014887994238259270716355378069726760953320025828158
tmp =  864078778078609835167779565982540757684070450697854309005171742813414963447462554999012718960925081621571487444725528982424037419052194840720949809891134854871222612682162490991065015935449289960707882463387
n =15911581555796798614711625288508309704791837516232122410440958830726078821069050404012820896260071751380436992710638364294658173571101596931605797509712839622479368850251206419748090059752427303611760004621378226431226983665746837779056271530181865648115862947527212787824629516204832313026456390047768174765687040950636530480549014401279054346098030395100387004111574278813749630986724706263655166289586230453975953773791945408589484679371854113457758157492241225180907090235116325034822993748409011554673180494306003272836905082473475046277554085737627846557240367696214081276345071055578169299060706794192776825039
#assert(pow(e1,ee1,n)==ce1)
#assert(pow(e2+tmp,ee2,n)==ce2)
#n  =  15911581555796798614711625288508309704791837516232122410440958830726078821069050404012820896260071751380436992710638364294658173571101596931605797509712839622479368850251206419748090059752427303611760004621378226431226983665746837779056271530181865648115862947527212787824629516204832313026456390047768174765687040950636530480549014401279054346098030395100387004111574278813749630986724706263655166289586230453975953773791945408589484679371854113457758157492241225180907090235116325034822993748409011554673180494306003272836905082473475046277554085737627846557240367696214081276345071055578169299060706794192776825039
def doit(ee,n,ce):
    k=0
    while True:
        x=ce+k*n
        if gmpy2.iroot(x,ee)[1]:
            return gmpy2.iroot(x,ee)[0]
        k=k+1
e1=doit(ee1,n,ce1)
e2=doit(ee2,n,ce2)-tmp
##小指数爆破
q1p=127587319253436643569312142058559706815497211661083866592534217079310497260365307426095661281103710042392775453866174657404985539066741684196020137840472950102380232067786400322600902938984916355631714439668326671310160916766472897536055371474076089779472372913037040153356437528808922911484049460342088834871
q1=q1p
##factordb爆破
q2 =  114401188227479584680884046151299704656920536168767132916589182357583461053336386996123783294932566567773695426689447410311969456458574731187512974868297092638677515283584994416382872450167046416573472658841627690987228528798356894803559278308702635288537653192098514966089168123710854679638671424978221959513
c1 =  262739975753930281690942784321252339035906196846340713237510382364557685379543498765074448825799342194332681181129770046075018122033421983227887719610112028230603166527303021036386350781414447347150383783816869784006598225583375458609586450854602862569022571672049158809874763812834044257419199631217527367046624888837755311215081173386523806086783266198390289097231168172692326653657393522561741947951887577156666663584249108899327053951891486355179939770150550995812478327735917006194574412518819299303783243886962455399783601229227718787081785391010424030509937403600351414176138124705168002288620664809270046124
c2 =  7395591129228876649030819616685821899204832684995757724924450812977470787822266387122334722132760470911599176362617225218345404468270014548817267727669872896838106451520392806497466576907063295603746660003188440170919490157250829308173310715318925771643105064882620746171266499859049038016902162599261409050907140823352990750298239508355767238575709803167676810456559665476121149766947851911064706646506705397091626648713684511780456955453552020460909638016134124590438425738826828694773960514221910109473941451471431637903182205738738109429736425025621308300895473186381826756650667842656050416299166317372707709596
#assert(c1==pow(flag,e1,p*q1))
#assert(c2==pow(flag,e2,p*q2))
phi1=(p-1)*(q1-1)
phi2=(p-1)*(q2-1)
xx1=gmpy2.gcd(e1,phi1)
xx2=gmpy2.gcd(e2,phi2)
d1=gmpy2.invert(e1//xx1,phi1)
d2=gmpy2.invert(e2//xx2,phi2)
nn=[]
aa=[]
nn.append(q1)
nn.append(q2)
a1=gmpy2.powmod(c1,d1,p*q1)%q1
a2=gmpy2.powmod(c2,d2,p*q2)%q2
aa.append(a1)
aa.append(a2)
last=exCRT_getequation(aa,nn)#最终方程组 aa=n^14%q1*q2
new_e=7
new_phi=(q1-1)*(q2-1)
new_d=gmpy2.invert(new_e,new_phi)
m_2=gmpy2.powmod(last[0],new_d,last[1])#特解m_2
flag=gmpy2.iroot(m_2,2)[0]
import binascii
print(str(binascii.unhexlify(hex(flag)[2:])).replace('de1ctf','flag'))

[MRCTF2020]Easy_RSA

知识点：简单的RSA魔改

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2021/6/1 14:47
# @Author  : upload
# @File    : RSA4.py
# @Software: PyCharm
import gmpy2
import sympy
from Crypto.Util.number import long_to_bytes

P_n =  14057332139537395701238463644827948204030576528558543283405966933509944444681257521108769303999679955371474546213196051386802936343092965202519504111238572269823072199039812208100301939365080328518578704076769147484922508482686658959347725753762078590928561862163337382463252361958145933210306431342748775024336556028267742021320891681762543660468484018686865891073110757394154024833552558863671537491089957038648328973790692356014778420333896705595252711514117478072828880198506187667924020260600124717243067420876363980538994101929437978668709128652587073901337310278665778299513763593234951137512120572797739181693
P_F_n =  14057332139537395701238463644827948204030576528558543283405966933509944444681257521108769303999679955371474546213196051386802936343092965202519504111238572269823072199039812208100301939365080328518578704076769147484922508482686658959347725753762078590928561862163337382463252361958145933210306431342748775024099427363967321110127562039879018616082926935567951378185280882426903064598376668106616694623540074057210432790309571018778281723710994930151635857933293394780142192586806292968028305922173313521186946635709194350912242693822450297748434301924950358561859804256788098033426537956252964976682327991427626735740
Q_n =  20714298338160449749545360743688018842877274054540852096459485283936802341271363766157976112525034004319938054034934880860956966585051684483662535780621673316774842614701726445870630109196016676725183412879870463432277629916669130494040403733295593655306104176367902352484367520262917943100467697540593925707162162616635533550262718808746254599456286578409187895171015796991910123804529825519519278388910483133813330902530160448972926096083990208243274548561238253002789474920730760001104048093295680593033327818821255300893423412192265814418546134015557579236219461780344469127987669565138930308525189944897421753947
Q_E_D =  100772079222298134586116156850742817855408127716962891929259868746672572602333918958075582671752493618259518286336122772703330183037221105058298653490794337885098499073583821832532798309513538383175233429533467348390389323225198805294950484802068148590902907221150968539067980432831310376368202773212266320112670699737501054831646286585142281419237572222713975646843555024731855688573834108711874406149540078253774349708158063055754932812675786123700768288048445326199880983717504538825498103789304873682191053050366806825802602658674268440844577955499368404019114913934477160428428662847012289516655310680119638600315228284298935201
Ciphertext =  40855937355228438525361161524441274634175356845950884889338630813182607485910094677909779126550263304194796000904384775495000943424070396334435810126536165332565417336797036611773382728344687175253081047586602838685027428292621557914514629024324794275772522013126464926990620140406412999485728750385876868115091735425577555027394033416643032644774339644654011686716639760512353355719065795222201167219831780961308225780478482467294410828543488412258764446494815238766185728454416691898859462532083437213793104823759147317613637881419787581920745151430394526712790608442960106537539121880514269830696341737507717448946962021

p = 118153578345562250550767057731385782963063734586321112579869747650001448473633860305142281504862521928246520876300707405515141444727550839066835195905927281903880307860942630322499106164191736174201506457157272220802515607939618476716593888428832962374494147723577980992661629254713116923690067827155668889571
q = 118975085954858660642562584152139261422493348532593400307960127317249511761542030451912561362687361053191375307180413931721355251895350936376781657674896801388806379750757264377396608174235075021854614328009897408824235800167369204203680938298803752964983358298299699273425596382268869237139724754214443556383
factor2 = 2021 * p + 2020 * q
if factor2 < 0:
    factor2 = (-1) * factor2
_P = sympy.nextprime(factor2)

pp = 120538849514661970159855851547577637711900368732462953774738483480759950867244867240401273864984981385806453735655967797329769252143125966966236767391995563418243748302685348336642872306042286401427581501609713577329945760930395130411743322595026287853073310150103535873078436896035943385067893062698858976291
qq = 171847486694659608706336923173786708071603689972942289760669690002615525263534483261477699540482615520223300780778172120221008417518590133753701145591943840552802072474293556608389677806415392384924913911677288126066245025731416399656855625839288752326267741979436855441260177305707529456715625062080892327017
factor22 = 2021 * pp - 2020 * qq
if factor22 < 0:
    factor22 = (-1) * factor22
_Q=sympy.nextprime(factor22)

fnnn = (_P-1)*(_Q-1)
base = 65537
_E = base
d = gmpy2.invert(_E, fnnn)
m =pow(Ciphertext,d,_P*_Q)
print(str(long_to_bytes(m)).replace('MRCTF','flag'))

[UTCTF2020]hill

知识点：希尔密码

a00, a01, a10, a11 = 13, 6, 3, 21
dic = 'abcdefghijklmnopqrstuvwxyz'
c = 'wznqcaduqopfkqnwofdbzgeu'
m = ''
for i in xrange(0, len(c), 2):
    a, b = dic.index(c[i]), dic.index(c[i+1])
    a0, b0 = (a00*a+a01*b)%26, (a10*a+a11*b)%26
    m += dic[a0]
    m += dic[b0]
print m