2024DASCTF暑期—Crypto复现

Zsm published on 2024-12-12 included in CTF

前言

感觉这是打的最有意思的一次比赛了，当时在火车站打开电脑写的，秒了一题，然后上火车了，一路很颠就没有再写了，复现一下前三题

题目

complex_enc

task

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31


from Crypto.Util.number import *
import random
from secret import flag


def GET_KEY(n):
    sum=2
    key=[1]
    for i in range(n):
        r=random.randint(0,1)
        x=sum+random.randint(0,n)*r
        key.append(x)
        sum+=x
    return key

def enc(m,k):
    cipher_list = []
    for i in range(len(m)):
        if m[i] == 1:
            cipher_list.append(m[i] * k[i])
    cipher = sum(cipher_list)
    return cipher

m=bytes_to_long(flag)
m = [int(bit) for byte in flag for bit in format(byte, '08b')]
key=GET_KEY(len(m))
c=enc(m,key)

with open('output.txt', 'w') as f:
    f.write(str(c))
    f.write(str(key))

output

Basectf2024_crypto

Zsm published on 2024-12-12 included in CTF

week1の题目

你会算MD吗

task

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13


import hashlib

flag='BaseCTF{}'

output=[]
for i in flag:
    my_md5=hashlib.md5()
    my_md5.update(i.encode())
    output.append(my_md5.hexdigest())
print("output =",output)
'''
output = ['9d5ed678fe57bcca610140957afab571', '0cc175b9c0f1b6a831c399e269772661', '03c7c0ace395d80182db07ae2c30f034', 'e1671797c52e15f763380b45e841ec32', '0d61f8370cad1d412f80b84d143e1257', 'b9ece18c950afbfa6b0fdbfa4ff731d3', '800618943025315f869e4e1f09471012', 'f95b70fdc3088560732a5ac135644506', '0cc175b9c0f1b6a831c399e269772661', 'a87ff679a2f3e71d9181a67b7542122c', '92eb5ffee6ae2fec3ad71c777531578f', '8fa14cdd754f91cc6554c9e71929cce7', 'a87ff679a2f3e71d9181a67b7542122c', 'eccbc87e4b5ce2fe28308fd9f2a7baf3', '0cc175b9c0f1b6a831c399e269772661', 'e4da3b7fbbce2345d7772b0674a318d5', '336d5ebc5436534e61d16e63ddfca327', 'eccbc87e4b5ce2fe28308fd9f2a7baf3', '8fa14cdd754f91cc6554c9e71929cce7', '8fa14cdd754f91cc6554c9e71929cce7', '45c48cce2e2d7fbdea1afc51c7c6ad26', '336d5ebc5436534e61d16e63ddfca327', 'a87ff679a2f3e71d9181a67b7542122c', '8f14e45fceea167a5a36dedd4bea2543', '1679091c5a880faf6fb5e6087eb1b2dc', 'a87ff679a2f3e71d9181a67b7542122c', '336d5ebc5436534e61d16e63ddfca327', '92eb5ffee6ae2fec3ad71c777531578f', '8277e0910d750195b448797616e091ad', '0cc175b9c0f1b6a831c399e269772661', 'c81e728d9d4c2f636f067f89cc14862c', '336d5ebc5436534e61d16e63ddfca327', '0cc175b9c0f1b6a831c399e269772661', '8fa14cdd754f91cc6554c9e71929cce7', 'c9f0f895fb98ab9159f51fd0297e236d', 'e1671797c52e15f763380b45e841ec32', 'e1671797c52e15f763380b45e841ec32', 'a87ff679a2f3e71d9181a67b7542122c', '8277e0910d750195b448797616e091ad', '92eb5ffee6ae2fec3ad71c777531578f', '45c48cce2e2d7fbdea1afc51c7c6ad26', '0cc175b9c0f1b6a831c399e269772661', 'c9f0f895fb98ab9159f51fd0297e236d', '0cc175b9c0f1b6a831c399e269772661', 'cbb184dd8e05c9709e5dcaedaa0495cf']
'''

思路：直接利用哈希表去找字符串就行了 exp

isctf2024_crypto

Zsm published on 2024-12-12 included in CTF

前言

本来以为社会组没有奖，就只写了crypto和一部分misc，结果结束了又说前十有奖，呜呜呜

题目

我和小蓝鲨的秘密

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25


from PIL import Image
from Crypto.Util.number import bytes_to_long, long_to_bytes
import numpy as np

n = 29869349657224745144762606999
e = 65537

original_image_path = "flag.jpg"
img = Image.open(original_image_path)
img = img.convert("RGB")

img_array = np.array(img)
h, w, _ = img_array.shape

encrypted_array = np.zeros((h, w, 3), dtype=object)
for i in range(h):
    for j in range(w):
        r, g, b = int(img_array[i, j, 0]), int(img_array[i, j, 1]), int(img_array[i, j, 2])

        encrypted_array[i, j, 0] = pow(r, e, n)
        encrypted_array[i, j, 1] = pow(g, e, n)
        encrypted_array[i, j, 2] = pow(b, e, n)

np.save("encrypted_image.npy", encrypted_array)
print("图片已加密并保存为 encrypted_image.npy")

思路：就是很简单图片rsa，可以直接，遍历，n很小，可以直接分解 exp

moectf2024_crypto

Zsm published on 2024-12-12 included in CTF

题目

现代密码学指北

task:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18


from Crypto.Util.number import bytes_to_long, getPrime
from secret import flag
p = getPrime(128)
q = getPrime(128)
n = p*q
e = 65537
m = bytes_to_long(flag)
c = pow(m, e, n)
print(f"n = {n}")
print(f"p = {p}")
print(f"q = {q}")
print(f"c = {c}")
'''
n = 40600296529065757616876034307502386207424439675894291036278463517602256790833
p = 197380555956482914197022424175976066223
q = 205695522197318297682903544013139543071
c = 36450632910287169149899281952743051320560762944710752155402435752196566406306
'''