import numpy as np import matplotlib.pyplot as plt
a = "1111111010010101010110111111110000010100011000111101000001101110100110100101000010111011011101010000110001000101110110111010110100100101101011101100000100100111011001010000011111111010101010101010111111100000000001111001000000000000110011010000110101010011011110001110110101101011000011110000110100101110010010111111000100001101000011010000100000100100101101001011010010101111010011110100111101001110000011110100101101001011010010001110011101101111011001110000001001010101001100011001001001001110111111111111011101101001010111001101110111001001110111000001111100011001000010100110110011111011010100111110010001100100011110011001100000000001001010101101010111001111111010110001101000101001010000011100111110010001000100101110101010011111111110111101011101000000010000000000111010111011110001010110111001110100000101110011101101110011101111111"
w = int(len(a)**0.5)
a = np.array([int(i) for i in a])
a.resize((w, w))
plt.imshow(a,'gray') plt.show()
扫码得到
1
033yia8rqea1921ca61/systemlockdown
补上比赛附件下载的前缀http://175.178.148.197下载，白盒pwn，研究一下
1
for (i = 0; i <= 6; i++) {
发现多循环了一次，可以输入七位，且每一位和前面的必须相等，随便爆破一下
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import string from pwn.toplevel import process
for i in string.digits: p = process("./a") p.sendline(i*7) if'Congurations'in (ret := p.recvall().decode()): print(i) break p.close()
Key Hashes Part 1: https://pastebin.com/QZ7QBmmd Key Hashes Part 2: https://pastebin.com/TUNVRVvk Key Hashes Part 3: https://pastebin.com/rTqtad96
有些加重的字体还有几封加密邮件，
直接扔进cmd5里面查（查了我20块钱，好贵。。。
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2fae32629d4ef4fc6341f1751b405e45 Security a498382929241d9ba043e11a272750af is e2a57aade7228356cccee547d2b6f91f too cdfaa4a64a0b1ef0fb69b98aed692fd6 often 9666ea9113697bd46fa8529a485de7c6 merely 1dcc6dfa679cb0654dea2f3bfa0da289 an 833a2eda1f339f7ba666c77cb4280150 illus 9f55995571880efd0b338989fec42d74 ion, ....
