BUUCTF crypto做题记录（7）新手向

一、Dangerous RSA

得到的密文如下

首先，我们对n进行大数分解看行不行。

其次，我们可看一下数的特征（除了一些基础题，一般情况下n都是分解不了的，应该首先观察一下数据特征，我很久没做RSA了，有点生疏），可以看到e特别小，所以我们可以对e展开一下低加密指数爆破攻击。

n=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
e=0x3
c=0x10652cdfaa6b63f6d7bd1109da08181e500e5643f5b240a9024bfa84d5f2cac9310562978347bb232d63e7289283871efab83d84ff5a7b64a94a79d34cfbd4ef121723ba1f663e514f83f6f01492b4e13e1bb4296d96ea5a353d3bf2edd2f449c03c4a3e995237985a596908adc741f32365
import gmpy2
from Crypto.Util.number import *

for i in range(0,10):
    x=c+i*n
    m=gmpy2.iroot(x, 3)
    m=long_to_bytes(m[0])
    if b'flag' in m:
        print(m)
        print(i)
        break

得到的答案为 flag{25df8caf006ee5db94d48144c33b2c3b}

二、[HDCTF2019]basic rsa

得到如下代码：

import gmpy2
from Crypto.Util.number import *
from binascii import a2b_hex,b2a_hex

flag = "*****************"

p = 262248800182277040650192055439906580479
q = 262854994239322828547925595487519915551

e = 65533
n = p*q


c = pow(int(b2a_hex(flag),16),e,n)

print c

# 27565231154623519221597938803435789010285480123476977081867877272451638645710

p和q都给出来了，就是一个直接的解密了

import gmpy2
from Crypto.Util.number import *
from binascii import a2b_hex,b2a_hex

flag = "*****************"

p = 262248800182277040650192055439906580479
q = 262854994239322828547925595487519915551

e = 65533
n = p*q


# c = pow(int(b2a_hex(flag),16),e,n)

c=27565231154623519221597938803435789010285480123476977081867877272451638645710
d=inverse(e,(p-1)*(q-1))
m=gmpy2.powmod(c, d, n)

print(long2str(m))
# 27565231154623519221597938803435789010285480123476977081867877272451638645710

得到答案：flag{B4by_Rs4}