2022-01-17

silent_peeper

（来源：暗泉杯）

from Crypto.Util.number import *
from Crypto.Cipher import AES
import binasciiflag = "flag{XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX}"
bs = AES.block_size
pad = lambda s: s + (bs - len(s) % bs) * chr(bs - len(s) % bs)p = 174807157365465092731323561678522236549173502913317875393564963123330281052524687450754910240009920154525635325209526987433833785499384204819179549544106498491589834195860008906875039418684191252537604123129659746721614402346449135195832955793815709136053198207712511838753919608894095907732099313139446299843
g = 41899070570517490692126143234857256603477072005476801644745865627893958675820606802876173648371028044404957307185876963051595214534530501331532626624926034521316281025445575243636197258111995884364277423716373007329751928366973332463469104730271236078593527144954324116802080620822212777139186990364810367977
a = getRandomNBitInteger(40)
b = getRandomNBitInteger(40)
A = pow(g, a, p)
B = pow(g, b, p)
assert pow(A, b, p) == pow(B, a, p)
key = pow(A, b ,p)
key = long_to_bytes(key)[:16]
cipher = AES.new(key, AES.MODE_ECB)
ciphertext = cipher.encrypt(pad(flag))with open('cipher', 'w') as f:f.write("A, B = {}\n".format(str((A, B))))f.write("ciphertext = {}\n".format(binascii.hexlify(ciphertext)))f.close()

A, B = (142989488568573584455487421652639325256968267580899511353325709765313839485530879575182195391847106611058986646758739505820350416810754259522949402428485456431884223161690132385605038767582431070875138678612435983425500273038807582069763455994486365993366499478412783220052753597397455113133312907456163112016L, 16631700400183329608792112442038543911563829699195024819408410612490671355739728510944167852170853457830111233224257622677296345757516691802411264928943809622556723315310581871447325139349242754287009766402650270061476954875266747743058962546605854650101122523183742112737784691464177427011570888040416109544L)
ciphertext = ed5c68ebb65aa3a13afb259cf3984ce60bdc54b7ef918b850745df850cf4c450b02216c0c6e67ed501a17e516496cd6c

这里用到了BSGS算法（大步小步算法）

它是用来解决这样一类问题
yx=z(mod p)，给定y,z,p>=1求解x
普通的BSGS只能用来解决gcd(y,p)=1的情况
设x=a∗m+b,m=⌈p–√⌉,a∈[0,m),b∈[0,m)
那么ya∗m=z∗y−b(mod p)
怎么求解，为了方便，设x=a∗m−b
那么ya∗m=z∗yb(mod p),a∈(0,m+1]
直接暴力辣，把右边的b枚举[0,m)，算出z∗yb(mod p)，哈希存起来
然后左边a枚举(0,m+1]，算出ya∗m(mod p)查表就行了
然后不知道为什么要用exgcd，只会map…

用sage解

sage: p = 17480715736546509273132356167852223654917350291331787539356496312333028105252468745075491024000992015452563532
....: 520952698743383378549938420481917954954410649849158983419586000890687503941868419125253760412312965974672161440234
....: 6449135195832955793815709136053198207712511838753919608894095907732099313139446299843
....: g = 41899070570517490692126143234857256603477072005476801644745865627893958675820606802876173648371028044404957307
....: 185876963051595214534530501331532626624926034521316281025445575243636197258111995884364277423716373007329751928366
....: 973332463469104730271236078593527144954324116802080620822212777139186990364810367977
....:
sage: bsgs(mod(g,p),mod(A,p),(0,1<<40))
822690494337
sage: bsgs(mod(g,p),mod(B,p),(0,1<<40))
621209248538
sage: key = pow(A, 621209248538 ,p)
sage: key
49490143664250726340234715933627573928019204778410313862054713655398194526581099674219755475997125892095025977920719640048704962181220475413665581922989858463397985369540020911109237604500080688916224884254427061443849735076051958183562833019840975221087968773423237208708556105725003184929141476854095400756

得到key之后转python

>>> from Crypto.Util.number import *
>>> from Crypto.Cipher import AES
>>> ciphertext = 0xed5c68ebb65aa3a13afb259cf3984ce60bdc54b7ef918b850745df850cf4c450b02216c0c6e67ed501a17e516496cd6c
>>> key = 49490143664250726340234715933627573928019204778410313862054713655398194526581099674219755475997125892095025977920719640048704962181220475413665581922989858463397985369540020911109237604500080688916224884254427061443849735076051958183562833019840975221087968773423237208708556105725003184929141476854095400756
>>> key = long_to_bytes(key)[:16]
>>> cipher = AES.new(key, AES.MODE_ECB)
>>> cipher.decrypt(int.to_bytes(ciphertext, ciphertext.bit_length()//8, 'big'))
b'flag{21384433-0dc7-413b-9d09-64cc97c99730}\x06\x06\x06\x06\x06\x06'

代码参考

Decrypt-It-easy

题目来源：攻防世界
题目描述：找到字符串在随机化之前.

先打开txt文件,

不知道要干什么，搜了别的师傅的解题，用IDA 打开rnd

找到main函数，F5查看代码

读a2[1]，写a2[2]

根据readme可以推断a2[1]就是那个png文件，a2[2]就是得到的ecrypt1.bin

又要用到png文件格式问题，类似beginners-luck
但也有很多不一样

依旧先根据png文件头进行爆破，其中时间是可以获取的

这里面就有c的函数，所以可以直接用C写就好了（代码参考）

#include <stdio.h>
#include <stdlib.h>int main(int argc, char *argv[]) {FILE *cipher = fopen(argv[1], "rb");FILE *plain = fopen(argv[2], "wb");unsigned int seed = atoi(argv[3]);int c;srand(seed);c = (fgetc(cipher) & 0xff) ^ (rand() & 0xff);while (!feof(cipher)) {fputc(c, plain);c = (fgetc(cipher) & 0xff) ^ (rand() & 0xff);}fclose(plain);fclose(cipher);
}

但是这个并不能直接运行，可以利用虚拟机一些命令来运行
最后得到

由于这是某次比赛的题目，我们还可以知道flag固定开头：SECCON{…}

所以就可以解了

N, B = 0xB86E78C811, 0xFFFEE
for i in range(32, 127):for j in range(32, 127):for k in range(32, 127):M = int(('N{' + chr(i) + chr(j) + chr(k)).encode('hex'), 16)if ((M * (M + B) % N) == 0x5FFA0AC1A2):print 'N{' + chr(i) + chr(j) + chr(k)
N, B = 0x7BD4071E55, 0xFEFEF
for i in range(32, 127):for j in range(32, 127):for k in range(32, 127):for l in range(32, 127):M = int((chr(i) + chr(j) + chr(k) + chr(l) + '}').encode('hex'), 16)if ((M * (M + B) % N) == 0x6008DDF867):print chr(i) + chr(j) + chr(k) + chr(l) + '}'

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2022-01-17

silent_peeper

Decrypt-It-easy

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