刷密码学刷到了这样的一道题
import random
from gmpy2 import next_prime
from flag import flag
def egcd(a, b):
x, y, u, v = 0, 1, 1, 0
while a != 0:
q, r = b//a, b % a
m, n = x-u*q, y-v*q
b, a, x, y, u, v = a, r, u, v, m, n
# print(m,n)
gcd = b
return gcd, x, y
def gen_keys(p, q):
e = 65537
n = p * q
phi = (p - 1) * (q - 1)
gcd, d, b = egcd(e, phi)
# Keys:((pub), (priv))
return ((e, n), (d, n))
def enc(key, p):
e, n = key
cipher = [int(pow(ord(char), e, n)) for char in p]
return cipher
def dec(pk, c):
key, n = pk
plain = [chr(pow(char, key, n)) for char in c]
return ''.join(plain)
p = next_prime(random.SystemRandom().getrandbits(512))
q = next_prime(random.SystemRandom().getrandbits(512))
flag_key=gen_keys(p, q)
print("Public key:")
print(flag_key[0])
flag_c=(enc(flag_key[0], flag))
print("Encrypted flag:")
print(flag_c)提供了两份output作为两道题 output我放在附件 实在有点大
题目读起来不是很困难 大概就是把flag中的每一个字符都作为m单独rsa，然后输出给我n和每一个m
试了一下factor结果拆不出来（肯定的吧！不然这么费劲出题干什么）
但是想了想这样先拆开再加密简直就是脱裤子放屁 反而更适合我爆破
import string
e = 65537
n = 28150970547901913019901824364390497053600856369839321617996700606130553862041378369018779003752433356889118526332329074054728613209407037089320809898343953157935211086135010436283805891893636870991411236307901650696221491790470635225076251966300189483160148297407974155121570252648252906976186499329924342873
cipher = ‘’‘这里值太长了 就是output的内容’‘’
chars = string.printable
flag = ''
for c in cipher:
found = False
for ch in chars:
if pow(ord(ch), e, n) == c:
flag += ch
found = True
break
if not found:
flag += '?'
print(flag)对着表很快就出来了nuaactf{F@k3_R5A_HahAHAHA}
第二份output明显阴间很多 没有给出n，但c足足有280,000bit 大概900字符？
而且题目描述上添油加醋地说
本题的答案不会有直接形如 flag{i_am_flag} 这样的答案，但是你会看到类似的内容 FLAG IS I AM FLAG 只需要提交 flag{I_AM_FLAG} 即可
言多必失啊.....思考了一会先写了这样一段脚本
def tongji(arr):
count = {}
for item in arr:
if item in count:
count[item] += 1
else:
count[item] = 1
return count
if __name__ == "__main__":
array=‘’‘依旧太多了写不下的output'''
result = tongji(array)
for value, count in result.items():
print(f"值 {value} 出现了 {count} 次")结果其中一个c重复了156次，另一个c重复了90次左右 疑似出题人写小作文 第一个是空格 第二个是e
分别代入进去 能求出kn和qn，求最大公因数就是n了，然后依旧爆破
c1=3454962493357935235804398048029156279039095550980789261896655337798360738722050425342188889752728340388460432418731237228933329641457678636310791152390769085756058697246856642532366705213972349136275749895187361949657257298558376286029052233131737322778610132480613456574001841996198028071886938400701639215
m1=32
e=65537
c2=5169002119818174391634714185903790851061018005401634820484671774278180792602838410733452791538037740062826920927698584515851903819438097478266878035790653024116345781217364538948047246739311236187915325479926636465158253396261869020517753466952653187629473271412557484931426586814990117865237871747512061719
m2=101
cipher=‘’‘你知道我这里要写什么’‘’
A = pow(m1, e)
y1 = A - c1
B = pow(m2,e)
y2 = B - c2
n = gmpy2.gcd(y1, y2)
flag = ''
for c in cipher:
found = False
for ch in range(32, 127):
if pow(ch, e, n) == c:
flag += chr(ch)
found = True
break
print(flag)运行结果是
don't starve is a marvelous survival game.the game opens with maxwell snidely commenting on the player's gaunt appearance and includes little further story.the game's setup is told further through its trailer: on a dark and stormy night, wilson appears to be getting nowhere in a chemistry experiment until he is startled by his radio speaking to him. it reveals that it has noticed his trouble and has secret knowledge of him. when he eagerly agrees, a flurry of equations and diagrams encircle him and fill his head. using white rats, a typewriter, and his blood among other tools and materials, wilson creates a giant machine. the radio commends his work and tells him to pull the machine's switch. he hesitates, but at the radio's insistence, he does so. the machine rattles violently, and a pair of ghostly arms whisk him into a different world while an apparition of maxwell cackles.** the flag is this is also fake rsa**
饥荒确实好玩 但下次别这么出题了
第一次在ctf中用统计频率写出题，好神奇（挠头）我会继续做下去的！