0x01 前言
这次比赛感觉打得并不好，密码就出了一题。。。。还是菜了，继续加油吧。
0x02 FlowerCipher
题目源码：
# python3
from secret import flag
import random
# flag = b'flag{%s}' % md5(something).hexdigest()
# note that md5 only have characters 'abcdef' and digits
def Flower(x, key):
flower = random.randint(0, 4096)
return x * (key ** 3 + flower)
flag = flag[5:-1]
rounds = len(flag)
L, R = 1, 0
for i in range(rounds):
L, R = R + Flower(L, flag[i]), L
print(L, R)
'''
15720197268945348388429429351303006925387388927292304717594511259390194100850889852747653387197205392431053069043632340374252629529419776874410817927770922310808632581666181899 139721425176294317602347104909475448503147767726747922243703132013053043430193232376860554749633894589164137720010858254771905261753520854314908256431590570426632742469003
'''$R_n = L_{n-1}，L_n = L_{n-2} + L_{n-1} * (key^3+flower)，对L_n模L_{n-1}, 由于L是递增的所以L_n mod L_{n-1} = L_{n-2}$
用下面的代码可以求出前一位的flag以及L和R，递归求解应该可以求出flag,冲！！
from gmpy2 import *
R = 139721425176294317602347104909475448503147767726747922243703132013053043430193232376860554749633894589164137720010858254771905261753520854314908256431590570426632742469003
L = 15720197268945348388429429351303006925387388927292304717594511259390194100850889852747653387197205392431053069043632340374252629529419776874410817927770922310808632581666181899
def decrypt(L, R):
L_pre2 = L % R
Flower = L - L_pre2
for i in range(0, 4096):
key = Flower // R - i
if iroot(key, 3)[1]:
print(int(iroot(key, 3)[0]), '(to ascii):', chr(int(iroot(key, 3)[0])), f'(i = {i})')
flag_bit = chr(int(iroot(key, 3)[0]))
if i == 4096:
print('oh no')
L = R
R = L_pre2
return flag_bit, L, R
flag = ''
while L != 1 and R != 0:
key, L, R = decrypt(L, R)
flag += str(key)
print('flag{' + str(flag) + '}')
#flag{052801c9bbcdd176e83d62fe66b708e3}0x03 最后
现在在作那道DSA的题，做出来之后应该也会写一篇wp，尽请期待。。。。。。