[网络协议]2022年春秋杯网络安全联赛春季赛勇者山峰密码部分wp

2022年春秋杯网络安全联赛春季赛勇者山峰部分wp

Crypto

bob’s enc

bob自己编写了一套加密算法，但是在加密的过程中，似乎混入了一些噪音，你能帮助他恢复出明文吗？

题目

#python2
from secret import *
import random

prime = 2141
print len(flag)
flag = map(ord, flag)
flag1 = flag[:21]
flag2 = flag[21:]
row = 64


def add(msg1, msg2):
    return [(x + y) % prime for x, y in zip(msg1, msg2)]


def multi(msg1, msg2):
    out = []
    for l in msg1:
        s = 0
        for x, y in zip(l, msg2):
            s += (x * y) % prime
            s %= prime
        out.append(s)
    return out


def genkey(leng):
    l = [[] for i in range(row)]
    for x in range(row):
        for i in range(leng):
            l[x].append(random.randint(0, 511))
    return l


key = genkey(len(flag1))
print key

cipher1 = multi(key, flag1)

print cipher1

cipher2 = multi(key, flag2)

noise = [random.randint(0, 6) for i in range(row)]
print add(noise, cipher2)

思路

将flag分为两部分（flag1和flag2），$ciphe{r}_1=multi(key,fla{g}_1),ciphe{r}_2=add(noise,multi(key,fla{g}_2))$,$ciphe{r}_1$是$key$和$fla{g}_1$相乘，可以看成矩阵相乘，那么反过来可求出$fla{g}_1(fla{g}_1=ke{y}^{?1}ciphe{r}_1)$；而$ciphe{r}_2$是在相乘基础上加上了噪音，在La佬博客上有这类的问题（LWE问题）。

博客下面有两个脚本，试了第一个发现并不行，换第二个可！

脚本

# sage
from sage.all import *
from sage.modules.free_module_integer import IntegerLattice
from random import randint
import sys
from itertools import starmap
from operator import mul

with open('out', 'r') as f:
    key = eval(f.readline())
    c1 = eval(f.readline())
    c2 = eval(f.readline())

prime = 2141

K = matrix(Zmod(prime), key[:21])
C1 = vector(Zmod(prime), c1[:21])
m1 = K.inverse() * C1
flag = ''
for i in m1:
    flag += chr(i)
print(flag)


# Babai's Nearest Plane algorithm
def Babai_closest_vector(M, G, target):
    small = target
    for _ in range(5):
        for i in reversed(range(M.nrows())):
            c = ((small * G[i]) / (G[i] * G[i])).round()
            small -= M[i] * c
    return target - small


m = 64
n = 21
q = prime

A_values = key
b_values = c2

A = matrix(ZZ, m + n, m)
for i in range(m):
    A[i, i] = q
for x in range(m):
    for y in range(n):
        A[m + y, x] = A_values[x][y]
lattice = IntegerLattice(A, lll_reduce=True)
# print("LLL done")
gram = lattice.reduced_basis.gram_schmidt()[0]
target = vector(ZZ, b_values)
res = Babai_closest_vector(lattice.reduced_basis, gram, target)
# print("Closest Vector: {}".format(res))

R = IntegerModRing(q)
M = Matrix(R, A_values)
ingredients = M.solve_right(res)

# print("Ingredients: {}".format(ingredients))

for i in ingredients:
    flag += chr(i)
print(flag)

# for row, b in zip(A_values, b_values):
#     effect = sum(starmap(mul, zip(map(int, ingredients), row))) % q
#     assert(abs(b - effect) < 2 ** 37)
#
# print("ok")

# flag{14e6f236-9eb9-46
# flag{14e6f236-9eb9-46fc-b636-4c54c3732e5f}

Train

题目内容：Alice发现了一辆小火车，她想要去成功登上这辆火车，你能帮帮她吗?

题目

from Crypto.Util.number import *
from hashlib import sha256
import socketserver
import signal
import string
import random
from secret import flag

banner = br'''
 .oooooo..o                             oooo  oooo  ooooooooooooo                     o8o              
d8P'    `Y8                             `888  `888  8'   888   `8                     `"'              
Y88bo.      ooo. .oo.  .oo.    .oooo.    888   888       888      oooo d8b  .oooo.   oooo  ooo. .oo.   
 `"Y8888o.  `888P"Y88bP"Y88b  `P  )88b   888   888       888      `888""8P `P  )88b  `888  `888P"Y88b  
     `"Y88b  888   888   888   .oP"888   888   888       888       888      .oP"888   888   888   888  
oo     .d8P  888   888   888  d8(  888   888   888       888       888     d8(  888   888   888   888  
8""88888P'  o888o o888o o888o `Y888""8o o888o o888o     o888o     d888b    `Y888""8o o888o o888o o888o 
'''

n0 = 30798082519452208630254982405300548841337042015746308462162479889627080155514391987610153873334549377764946092629701
g = 64146569863628228208271069055817252751116365290967978172021890038925428672043


def TrainHash(msg):
    n = n0
    msg = map(ord, msg)
    for i in msg:
        n = g * (n + i)
        n = n & (1 << 383)
    return n - 0xf5e33dabb114514


table = string.ascii_letters + string.digits

MENU = br'''
<OPTION>
'''


class Task(socketserver.BaseRequestHandler):
    def _recvall(self):
        BUFF_SIZE = 2048
        data = b''
        while True:
            part = self.request.recv(BUFF_SIZE)
            data += part
            if len(part) < BUFF_SIZE:
                break
        return data.strip()

    def send(self, msg, newline=True):
        try:
            if newline:
                msg += b'\n'
            self.request.sendall(msg)
        except:
            pass

    def recv(self, prompt=b'SERVER <INPUT>: '):
        self.send(prompt, newline=False)
        return self._recvall()

    def proof_of_work(self):
        proof = (''.join([random.choice(table) for _ in range(20)])).encode()
        sha = sha256(proof).hexdigest().encode()
        self.send(b"[+] sha256(XXXX+" + proof[4:] + b") == " + sha)
        XXXX = self.recv(prompt=b'[+] Plz Tell Me XXXX :')
        if len(XXXX) != 4 or sha256(XXXX + proof[4:]).hexdigest().encode() != sha:
            return False
        return sha.decode()

    def handle(self):
        signal.alarm(30)
        FirstBlockHash = self.proof_of_work()
        if not FirstBlockHash:
            self.request.close()
        self.send(banner)
        self.send(b"\nPlease give me 2 strings that are same when are hashed  =.=  ")
        string1 = self.recv().decode()
        string2 = self.recv().decode()

        if TrainHash(string1) == TrainHash(string2):
            self.send(b'\nJust do it!~ You can do more!')
            if string2.encode()[-50:] == string1.encode()[-50:]:
                self.send(flag)
        self.send(b"\nConnection has been closed  =.=  ")
        self.request.close()


class ThreadedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
    pass


class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
    pass


if __name__ == "__main__":
    HOST, PORT = '0.0.0.0', 10012
    print("HOST:POST " + HOST + ":" + str(PORT))
    server = ForkedServer((HOST, PORT), Task)
    server.allow_reuse_address = True
    server.serve_forever()

思路

分析个der，随便给两个一样的string就有flag了，送分题嗷

脚本

from pwn import *
import string
import hashlib
table = string.ascii_letters + string.digits
re = remote('101.200.198.40',35946)

re.recvuntil(b'[+] sha256(XXXX+')
x = re.recv(16)
re.recvuntil(b') == ')
y = re.recv(64)

flag = 0
for a in table:
    if flag:break
    for b in table:
        if flag: break
        for c in table:
            if flag: break
            for d in table:
                z = (a + b + c + d).encode()
                if hashlib.sha256(z + x).hexdigest() == y.decode():
                    flag = 1
                    re.recv()
                    re.sendline(z)
                    break
re.recvuntil(b'SERVER <INPUT>: ')
re.sendline(b'a')
re.recvuntil(b'SERVER <INPUT>: ')
re.sendline(b'a')
re.interactive()

.sha256(z + x).hexdigest() == y.decode():
                    flag = 1
                    re.recv()
                    re.sendline(z)
                    break
re.recvuntil(b'SERVER <INPUT>: ')
re.sendline(b'a')
re.recvuntil(b'SERVER <INPUT>: ')
re.sendline(b'a')
re.interactive()

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