作为一个刚接触CTF不到3个月的小白，抱着来学习的想法，参加了西湖论剑·2020中国杭州网络安全技能大赛，只做出来2道题目，发一下WriteUp。
一、[CRYPTO]BrokenSystems
介绍大意是，有人黑进了加密系统并进行了篡改，剩下的就交给你了。
提供了3个文件
1.加密算法脚本
from Crypto.PublicKey import RSA from Crypto.Cipher import PKCS1_OAEP from secret import flag import os rsa = RSA.generate(2048) public_key = rsa.publickey().exportKey() f=open("public.key","w") f.write(public_key.decode()) f.close() rsakey=RSA.importKey(open("public.key","r").read()) rsa = PKCS1_OAEP.new(rsakey) msg=rsa.encrypt(flag.encode()) f=open("message","wb") f.write(msg) f.close()
2.公钥public.key
-----BEGIN PUBLIC KEY----- MIICITANBgkqhkiG9w0BAQEFAAOCAg4AMIICCQKCAQEAwgdFIj/1uUss2EEhZvco iiHyGH4aQhRTkYyrA8gCU0USM+sb3CNjdEIoqoaUqMLLyDP4Dd9AgxpokBsjC4Pz 8P7Uty0LlCteld7ayNzABHoq+5DIHtctOtSbcvyL0NMWfd2qarUWfAWN82rxkLIW CFu9nWIfm9I6CT5OPZzDh7YnTywznIjhstkIrLM/TiDmR6vuBxSjzORkbolilLeB A9zJpNt+1oEWTG5sx/0zR24XSmxwcDeyUEkTZfnw63auq6B9svZi2IBIr5jIjHbG cQ25ZY1J/KDK8fXNmdwH8YhDK0j4VXEWitEPyCS3toK61sql0S/28EySeGtzirGb twKCAQAdLS8fFz+BzzaP7AbUDUf9kvhhXBLuwUFo8ohCeVK4z1pTj3C6M0G2oXOu gDdalrDThNlyKxkUn3iUc3Xgoz315pPtq9Xk1Ez/qeUl6gFPP6SZtfeymyGdkLiN pVquOghjczjXvtBW467Fdb5Wu95TSzVaLndX23rsqW541n8hUwt8PsJKxh+bR0qy gyIN2VRRNdBlpyTOL49E4y5GDu9fmVgAnFivWVGT135ywl8MsBUFuZPBNTKLEbUA 3KvJVckXf4Od0ENYbiWjEzXn1UN9yebNbU6+yyk34WAmwnkuF0X0Tu1UEb6qtV7Q kF25GYy9QxERvodGL0Y2njHRpGE/ -----END PUBLIC KEY-----
3.密文message
由于是二进制，这里转为BASE64贴出来，需要的自己转回去即可。
J6a4HIY3ilKlD3lwR7GRG5WCPANVLC4NV9UKv/jHw23jj6I6LhgFhuY3P3PAtMtoEA7Wu1r7vNH/xSaQGidEryCJ8SXdqH29AqfMfmEszf2KtZib2sl2fatU1gHsLCNhZIE5pSck7N+cjOBnDDDfVTsOI8nJhF8yEPkO7pyK02srFdDooySSzkhEumWWkblciuwDwb4A6uD0K7X45rsJQ0TN3/cZFTOnIGgf9RNWoDlwrFp8C/xkVVWbask14UzaqrfWYp3fGrORgH3DLn5RHUq7O+exk7AJgqr3qIlTDcQeEEnm860nt9FKQl+TFE8wJgIP0AH3wg1TeE7hr3kY4g==
解题思路：
首先从加密脚本确定是一个正常的RSA公钥加密，需要破解得到私钥才能解密。由于提供了公钥，我们能获得e和n的值，从而判断属于什么漏洞。
第一步，获取e和n：
from Crypto.PublicKey import RSA from rsa import PublicKey,transform, core from Crypto.Cipher import PKCS1_OAEP pubkey=open("public.key","r").read() rsakey = PublicKey.load_pkcs1_openssl_pem(pubkey) n=rsakey.n e=rsakey.e print(n) print(e)
观察打印出来的n和e：
24493816160588971749455534346389861269947121809901305744877671102517333076424951483888863597563544011725032585417200878377314372325231470164799594965293350352923195632229495874587039720317200655351788887974047948082357232348155828924230567816817425104960545706688263839042183224681231800805037117758927837949941052360649778743187012198508745207332696876463490071925421229447425456903529626946628855874075846839745388326224970202749994059533831664092151570836853681204646481502222112116971464211748086292930029540995987019610460396057955900244074999111267618452967579699626655472948383601391620012180211885979095636919 3683191938452247871641914583009119792552938079110383367782698429399084083048335018186915282465581498846777124014232879019914546010406868697694661244001972931366227108140590201194336470785929194895915077935083045957890179080332615291089360169761324533970721460473221959270664692795701362942487885620152952927112838769014944652059440137350285198702402612151501564899791870051001152984815689187374906618917967106000628810361686645504356294175173529719443860140795170776862320812544438211122891112138748710073230404456268507750721647637959502454394140328030018450883598342764577147457231373121223878829298942493059211583
发现e很大，我们知道一般e取值为65537，当加密指数e取得过大时，会导致解密指数d较小，存在低解密指数攻击的可能性。我们套用破解脚本：
def rational_to_contfrac (x, y): \'\'\' Converts a rational x/y fraction into a list of partial quotients [a0, ..., an] \'\'\' a = x//y if a * y == x: return [a] else: pquotients = rational_to_contfrac(y, x - a * y) pquotients.insert(0, a) return pquotients def convergents_from_contfrac(frac): \'\'\' computes the list of convergents using the list of partial quotients \'\'\' convs = []; for i in range(len(frac)): convs.append(contfrac_to_rational(frac[0:i])) return convs def contfrac_to_rational (frac): \'\'\'Converts a finite continued fraction [a0, ..., an] to an x/y rational. \'\'\' if len(frac) == 0: return (0,1) elif len(frac) == 1: return (frac[0], 1) else: remainder = frac[1:len(frac)] (num, denom) = contfrac_to_rational(remainder) # fraction is now frac[0] + 1/(num/denom), which is # frac[0] + denom/num. return (frac[0] * num + denom, num) def egcd(a,b): \'\'\' Extended Euclidean Algorithm returns x, y, gcd(a,b) such that ax + by = gcd(a,b) \'\'\' u, u1 = 1, 0 v, v1 = 0, 1 while b: q = a // b u, u1 = u1, u - q * u1 v, v1 = v1, v - q * v1 a, b = b, a - q * b return u, v, a def gcd(a,b): \'\'\' 2.8 times faster than egcd(a,b)[2] \'\'\' a,b=(b,a) if a<b else (a,b) while b: a,b=b,a%b return a def modInverse(e,n): \'\'\' d such that de = 1 (mod n) e must be coprime to n this is assumed to be true \'\'\' return egcd(e,n)[0]%n def totient(p,q): \'\'\' Calculates the totient of pq \'\'\' return (p-1)*(q-1) def bitlength(x): \'\'\' Calculates the bitlength of x \'\'\' assert x >= 0 n = 0 while x > 0: n = n+1 x = x>>1 return n def isqrt(n): \'\'\' Calculates the integer square root for arbitrary large nonnegative integers \'\'\' if n < 0: raise ValueError(\'square root not defined for negative numbers\') if n == 0: return 0 a, b = divmod(bitlength(n), 2) x = 2**(a+b) while True: y = (x + n//x)//2 if y >= x: return x x = y def is_perfect_square(n): \'\'\' If n is a perfect square it returns sqrt(n), otherwise returns -1 \'\'\' h = n & 0xF; #last hexadecimal "digit" if h > 9: return -1 # return immediately in 6 cases out of 16. # Take advantage of Boolean short-circuit evaluation if ( h != 2 and h != 3 and h != 5 and h != 6 and h != 7 and h != 8 ): # take square root if you must t = isqrt(n) if t*t == n: return t else: return -1 return -1 def hack_RSA(e,n): frac = rational_to_contfrac(e, n) convergents = convergents_from_contfrac(frac) for (k,d) in convergents: #check if d is actually the key if k!=0 and (e*d-1)%k == 0: phi = (e*d-1)//k s = n - phi + 1 # check if the equation x^2 - s*x + n = 0 # has integer roots discr = s*s - 4*n if(discr>=0): t = is_perfect_square(discr) if t!=-1 and (s+t)%2==0: print("\nHacked!") return d def main(): n = 24493816160588971749455534346389861269947121809901305744877671102517333076424951483888863597563544011725032585417200878377314372325231470164799594965293350352923195632229495874587039720317200655351788887974047948082357232348155828924230567816817425104960545706688263839042183224681231800805037117758927837949941052360649778743187012198508745207332696876463490071925421229447425456903529626946628855874075846839745388326224970202749994059533831664092151570836853681204646481502222112116971464211748086292930029540995987019610460396057955900244074999111267618452967579699626655472948383601391620012180211885979095636919 e = 3683191938452247871641914583009119792552938079110383367782698429399084083048335018186915282465581498846777124014232879019914546010406868697694661244001972931366227108140590201194336470785929194895915077935083045957890179080332615291089360169761324533970721460473221959270664692795701362942487885620152952927112838769014944652059440137350285198702402612151501564899791870051001152984815689187374906618917967106000628810361686645504356294175173529719443860140795170776862320812544438211122891112138748710073230404456268507750721647637959502454394140328030018450883598342764577147457231373121223878829298942493059211583 d=hack_RSA(e,n) print ("d=") print (d) if __name__ == \'__main__\': main()
输出结果为：
Hacked!
d=
1779217788383673416690068487595062922771414230914791138743960472798057054853883175313487137767631446949382388070798609545617543049566741624609996040273727
这样就得到了d。如果直接用pow(c,d,n)尝试计算原文，会发现输出是乱码，琢磨了半天发现，我们应当构造RSA私钥进行解密：
from Crypto.PublicKey import RSA from rsa import PublicKey,transform, core from Crypto.Cipher import PKCS1_OAEP private_key = RSA.construct((n, e, d)) rsa = PKCS1_OAEP.new(private_key) m=rsa.decrypt(long_to_bytes(c)) print(m)
这样就解出了原文：
DASCTF{ce02347b86167f2d3519251b9a8a5ba8}
二、[Misc]指鹿为马
题目给了一个地址和端口，没有附件。用nc连上去一看，长这样：
--------------------------------------------------------------------------------------------------------------------------------------
____ __ _ _ _ _ _ _ _
| __ \ / _| | | | | | | | | | | | | | |
| |__) |___| |_ ___ _ __ | |_ ___ | |_| |__ ___ __| | ___ ___ _ __ __ _ ___ | |_| |__ ___ | |__ ___ _ __ ___ ___
| _ // _ \ _/ _ \ \'__| | __/ _ \ | __| \'_ \ / _ \ / _` |/ _ \/ _ \ \'__| / _` / __| | __| \'_ \ / _ \ | \'_ \ / _ \| \'__/ __|/ _ \
| | \ \ __/ || __/ | | || (_) | | |_| | | | __/ | (_| | __/ __/ | | (_| \__ \ | |_| | | | __/ | | | | (_) | | \__ \ __/
|_| \_\___|_| \___|_| \__\___/ \__|_| |_|\___| \__,_|\___|\___|_| \__,_|___/ \__|_| |_|\___| |_| |_|\___/|_| |___/\___|
--------------------------------------------------------------------------------------------------------------------------------------
1.show source code
2.give me the source pictures
3.upload picture
4.exit
Refer to the deer as the horse，好一个指鹿为马。
输入1，获得源文件代码：
import numpy as np from PIL import Image import math import operator import os import time import base64 import random def load_horse(): data = [] p = Image.open(\'./horse.png\').convert(\'L\') p = np.array(p).reshape(-1) p = np.append(p,0) data.append(p) return np.array(data) def load_deer(): data = [] p = Image.open(\'./deer.png\').convert(\'L\') p = np.array(p).reshape(-1) p = np.append(p,1) data.append(p) return np.array(data) def load_test(pic): data = [] p = Image.open(pic).convert(\'L\') p = np.array(p).reshape(-1) p = np.append(p,1) data.append(p) return np.array(data) def euclideanDistance(instance1, instance2, length): distance = 0 for x in range(length): distance += pow((instance1[x] - instance2[x]), 2) return math.sqrt(distance) def getNeighbors(trainingSet, testInstance, k):#k===1 distances = [] length = len(testInstance) - 1 for x in range(len(trainingSet)): dist = euclideanDistance(testInstance, trainingSet[x], length) distances.append((trainingSet[x], dist)) distances.sort(key=operator.itemgetter(1)) print(\'Eud: \', distances) neighbors = [] for x in range(k):#(0,1) neighbors.append(distances[x][0]) return neighbors def getResponse(neighbors): classVotes = {} for x in range(len(neighbors)): response = neighbors[x][-1] if response in classVotes: classVotes[response] += 1 else: classVotes[response] = 1 sortedVotes = sorted(classVotes.items(), key=operator.itemgetter(1), reverse=True) return sortedVotes[0][0] def getAccuracy(testSet, predictions): correct = 0 for x in range(len(testSet)): if testSet[x][-1] == predictions[x]: correct += 1 return (correct / float(len(testSet))) * 100.0 def check(pic): source_p = Image.open(\'deer.png\') try: c_p = Image.open(pic) except: print("Please upload right picture.") exit() diff_pixel = 0 a, b = source_p.size if c_p.size[0] != a and c_p.size[1] != b: print("Please upload right picture size("+str(a)+\',\'+str(b)+\')\') exit() for y in range(b): for x in range(a): diff_pixel += abs(source_p.getpixel((x, y)) - c_p.getpixel((x, y))) print(diff_pixel) return diff_pixel def main(): while 1: print(\'-\' * 134) print(\'\'\' ____ __ _ _ _ _ _ _ _ | __ \ / _| | | | | | | | | | | | | | | | |__) |___| |_ ___ _ __ | |_ ___ | |_| |__ ___ __| | ___ ___ _ __ __ _ ___ | |_| |__ ___ | |__ ___ _ __ ___ ___ | _ // _ \ _/ _ \ \'__| | __/ _ \ | __| \'_ \ / _ \ / _` |/ _ \/ _ \ \'__| / _` / __| | __| \'_ \ / _ \ | \'_ \ / _ \| \'__/ __|/ _ \\ | | \ \ __/ || __/ | | || (_) | | |_| | | | __/ | (_| | __/ __/ | | (_| \__ \ | |_| | | | __/ | | | | (_) | | \__ \ __/ |_| \_\___|_| \___|_| \__\___/ \__|_| |_|\___| \__,_|\___|\___|_| \__,_|___/ \__|_| |_|\___| |_| |_|\___/|_| |___/\___| \'\'\') print(\'-\'*134) print(\'\t1.show source code\') print(\'\t2.give me the source pictures\') print(\'\t3.upload picture\') print(\'\t4.exit\') choose = input(\'>\') print(choose) if choose == \'1\': w = open(\'run.py\',\'r\') print(w.read()) continue elif choose == \'2\': print(\'this is horse`s picture:\') h = base64.b64encode(open(\'horse.png\',\'rb\').read()) print(h.decode()) print(\'-\'*134) print(\'this is deer`s picture:\') d = base64.b64encode(open(\'deer.png\', \'rb\').read()) print(d.decode()) continue elif choose == \'4\': break elif choose == \'3\': print(\'Please input your deer picture`s base64(Preferably in png format)\') pic = input(\'>\') try: pic = base64.b64decode(pic) except: exit() if b"<?php" in pic or b\'eval\' in pic: print("Hacker!!This is not WEB,It`s Just a misc!!!") exit() salt = str(random.getrandbits(15)) pic_name = \'tmp_\'+salt+\'.png\' tmp_pic = open(pic_name,\'wb\') tmp_pic.write(pic) tmp_pic.close() #跟deer要相似 if check(pic_name)>=100000: print(\'Don`t give me the horse source picture!!!\') os.remove(pic_name) break ma = load_horse() print(\'ma:\') print(ma) lu = load_deer() print(\'lu:\') print(lu) k = 1 trainingSet = np.append(ma, lu).reshape(2, 5185) print(\'train:\', trainingSet) testSet = load_test(pic_name) print(\'test:\', testSet, len(testSet[0])) neighbors = getNeighbors(trainingSet, testSet[0], k) print(\'neighbors:\', neighbors) result = getResponse(neighbors) if repr(result) == \'0\': os.system(\'clear\') print(\'Yes,I want this horse like deer,here is your flag encoded by base64\') flag = base64.b64encode(open(\'flag\',\'rb\').read()) print(flag.decode()) os.remove(pic_name) break else: print(\'I want horse but not deer!!!\') os.remove(pic_name) break else: print(\'wrong choose!!!\') break exit() if __name__==\'__main__\': main()
输入2，获得图片的BASE64值：
this is horse`s picture: 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 -------------------------------------------------------------------------------------------------------------------------------------- this is deer`s picture: 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 --------------------------------------------------------------------------------------------------------------------------------------
输入3，提示输入图片：Please input your deer picture`s base64(Preferably in png format)
解题思路：
阅读代码，发现需要用户上传的图片符合一定要求，才会给出flag。这个要求就是，看起来是鹿，却又更像马-_-||
具体来说，就是以check函数计算出的距离来看，与鹿的图片差距值小于一定阈值（100000），而用euclideanDistance（欧氏距离）作为依据又更接近马的图片。
乍一看，这有可能吗？我们来分析一下。
我们把问题简化，假如图片只有2个像素，IMG鹿=(x1, y1), IMG马=(x2, y2)，用户上传图片IMG_USER=(xt, yt)，则要求：
abs(x1-xt) + abs(y1-yt) < T
且：
d1 = sqrt((x1 - xt)^2 + (y1 - yt)^2)
d2 = sqrt((x2 - xt)^2 + (y2 - yt)^2)
min(d1, d2) == d2
在坐标系中表示出来，就是这样：
我们需要找到一个点User(xt, yt)，落在图示中的位置，这就是答案。
而题目中的向量是图片的灰度矩阵拉平后的值，有5184维。我们无法求解出这个目标区域的解析解，只需要找到一个符合的向量就可以了。
我是通过在鹿的图片中替换一部分像素为马的像素来求解，脚本如下：
from PIL import Image import numpy as np from PIL import Image import math import os import time import base64 import random def euclideanDistance(instance1, instance2, length): distance = 0 for x in range(length): distance += pow((instance1[x] - instance2[x]), 2) return math.sqrt(distance) def check(p1, p2): diff_pixel=0 for x in range(len(p1)): diff_pixel += abs(p1[x] - p2[x]) #print(diff_pixel) return diff_pixel p1 = Image.open(\'./horse.png\').convert(\'L\') p1 = np.array(p1).reshape(-1) print(\'horse\', p1) p2 = Image.open(\'./deer.png\').convert(\'L\') p2 = np.array(p2).reshape(-1) print(\'deer\', p2) test = np.array(p2).reshape(-1) for i in range(3000): test[i] = p1[i] eud1 = euclideanDistance(p1, test, 5184) print(\'Eu1: \', eud1) eud2 = euclideanDistance(p2, test, 5184) print(\'Eu2: \', eud2) dif = check(test, p2) print(\'Diff:\', dif) if(eud1 < eud2): print(\'Eud Good\') if dif < 100000: print(\'Diff Good\') img = test.reshape(72, 72) #Image.save(img, \'test.png\') img = np.uint8(img) im = Image.fromarray(img) im.save("out.png", \'png\') with open("out.png", \'rb\') as f: base64_data = base64.b64encode(f.read()) s = base64_data.decode() print(s)
得到的图片BASE64值输入到服务器后，返回了另一段BASE64，转为PNG图片后，得到flag：
由于本人水平有限，只做出来这两道题目，与道友们共同学习多多交流，如有纰漏欢迎指正。