iscc2018-Reverse-writeup

RSA256

解析公钥

yafu质因数分解

p=325045504186436346209877301320131277983

q=302825536744096741518546212761194311477

e=65537

生成

 import math
 import sys
 from Crypto.PublicKey import RSA

 keypair = RSA.generate(1024)

 keypair.p = 325045504186436346209877301320131277983
 keypair.q = 302825536744096741518546212761194311477
 keypair.e = 65537
 keypair.n = keypair.p * keypair.q
 q = long((keypair.p-1) * (keypair.q-1))
 i = 1
 while True:
     x=(q*i)+1
     if (x % keypair.e == 0):
         keypair.d = x / keypair.e
         break
     i += 1
 print keypair.d #
 private = open('private.pem','w')
 private.write(keypair.exportKey())
 private.close()

解密脚本

 #coding=utf-8
 import rsa,base64

 with open('private.pem','r') as f:
     privkey = rsa.PrivateKey.load_pkcs1(f.read().encode())

 f1=open('encrypted.message1','r')
 f2=open('encrypted.message2','r')
 f3=open('encrypted.message3','r')
 a=f1.read()
 message1 = rsa.decrypt(a, privkey).decode().rstrip()
 a=f2.read()
 message2 = rsa.decrypt(a, privkey).decode().rstrip()
 a=f3.read()
 message3 = rsa.decrypt(a, privkey).decode().rstrip()
 print message1+message2+message3 #flag{3b6d3806-4b2b-11e7-95a0-000c29d7e93d}

leftleftrightright

peid查壳

手动脱壳

ida查看伪代码

还原flag：Flag{this_was_simple_isnt_it}

My math is bad

elf64，ida查看，题目要求为两个四元方程组求解

 from sympy import *
 a=Symbol('a')
 b=Symbol('b')
 c=Symbol('c')
 d=Symbol('d')
 e=Symbol('e')
 f=Symbol('f')
 g=Symbol('g')
 h=Symbol('h')
 print solve([
 a*b-c*d-2652042832920173142,
 3*c+4*d-b-2*a-397958918,
 3*d*a-b*c-3345692380376715070,
 27*b+a-11*d-c-40179413815
 ],[a,b,c,d])[0]
 print solve([
 22*e-f-g+39*h-61799700179,
 45*g-45*f+e+h-48753725643,
 35*e+41*f-g-h-59322698861,
 e-f+36*g-13*h-51664230587
 ],[e,f,g,h])

 a=1869639009
 b=1801073242
 c=829124174
 d=862734414
 e=811816014
 f=828593230
 g=1867395930
 h=1195788129
 l=[a,b,c,d,e,f,g,h]
 s=''
 for i in l:
     c=''
     for j in range(4):
         c+=chr(int(hex(i)[2+j*2:4+j*2],16))
     s+=c[::-1]
 print s #ampoZ2ZkNnk1NHl3NTc0NTc1Z3NoaGFG

obfuscation and encode

import string,base64
aphla='FeVYKw6a0lDIOsnZQ5EAf2MvjS1GUiLWPTtH4JqRgu3dbC8hrcNo9/mxzpXBky7+'
base='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
s='lUFBuT7hADvItXEGn7KgTEjqw8U5VQUq'
m=[2 ,2, 4, -5, 1 ,1, 3, -3, -1, -2, -3, 4, -1,0, -2, 2]
s1=''
for c in s:
    s1+=base[aphla.index(c)]
s1=base64.b64decode(s1)
for i in string.printable:
    for j in string.printable:
        for k in string.printable:
            for l in string.printable:
                s=i+j+k+l
                s0=''
                for n in range(4):
                    v10=0
                    for o in range(4):
                        v10+=ord(s[o])*m[4*n+o]
                    s0+=chr(v10%256)
                if s0==s1[0:4]:
                    print s,1
                if s0==s1[4:8]:
                    print s,2
                if s0==s1[8:12]:
                    print s,3
                if s0==s1[12:16]:
                    print s,4
                if s0==s1[16:20]:
                    print s,5
                if s0==s1[20:24]:
                    print s,6
'''
s3='flag{dO_y0U_KNoW_0IlVm?}'