crypto第二面前几题

1.Dangerous RSA

解压缩之后：

#n: 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
#e: 0x3
#c:0x10652cdfaa6b63f6d7bd1109da08181e500e5643f5b240a9024bfa84d5f2cac9310562978347bb232d63e7289283871efab83d84ff5a7b64a94a79d34cfbd4ef121723ba1f663e514f83f6f01492b4e13e1bb4296d96ea5a353d3bf2edd2f449c03c4a3e995237985a596908adc741f32365
so,how to get the message?

学习了一下发现这是低加密指数攻击：
所谓低加密指数指的就是e非常小的情况下，通常为3。
这种题目通常有两种类型，一种直接爆破，另外一种是低指数广播攻击。

假设e=3, e很小，但是n很大。
回顾RSA加密公式： C=M^e % n (C密文，M明文)

抄来的代码用一下发现

'''
当M^e < n 时，
 C = M^e ，所以对C开方就能得到M
'''
from gmpy2 import iroot
import libnum
n = 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

c = 0x10652cdfaa6b63f6d7bd1109da08181e500e5643f5b240a9024bfa84d5f2cac9310562978347bb232d63e7289283871efab83d84ff5a7b64a94a79d34cfbd4ef121723ba1f663e514f83f6f01492b4e13e1bb4296d96ea5a353d3bf2edd2f449c03c4a3e995237985a596908adc741f32365

k = 0
while 1:
    res=iroot(c+k*n,3)
    if(res[1]==True):
        print(libnum.n2s(int(res[0])))
        break
    k=k+1


'''

第二种写法
当M^e ＞ n 时，此时用爆破的方法
 假设我们　 Ｍ^e / n 商 k 余数为c，
 所以Ｍ^e  = k*n + C，对K进行爆破，只要k满足 k*n + C能够开方就可以
'''
'''
import gmpy2 
from libnum import*
n = 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
c = 0x10652cdfaa6b63f6d7bd1109da08181e500e5643f5b240a9024bfa84d5f2cac9310562978347bb232d63e7289283871efab83d84ff5a7b64a94a79d34cfbd4ef121723ba1f663e514f83f6f01492b4e13e1bb4296d96ea5a353d3bf2edd2f449c03c4a3e995237985a596908adc741f32365

i = 0
while 1:
    if(gmpy2.iroot(c+i*n,3)[1]==1):     #开根号
        print(gmpy2.iroot(c+i*n,3))
        break
    i=i+1

'''

得到

 

 

 2.Cipher

还能提示什么呢？公平的玩吧（密钥自己找） Dncnoqqfliqrpgeklwmppu 注意：得到的 flag 请包上 flag{} 提交, flag{小写字母}

 

 

 虽然看不明白，但是找到了一个网页试一下：

密钥也查了下资料发现：

 

 

 

 

 

 记得把大写换成小写

 

3.robomunication

音频文件，打开听一下，摩斯密码嘛，记录了一下：

bbbb b bpbb bpbb ppp bpp bbbb bp p

bb bbb p bbbb b pbp b pbpp bb p bb bbb pbbb ppp ppp bppb pbbb b b bppb

 

用.替换b，-替换p

 

 

 得到HELLOWHATISTHEKEYITISBOOPBEEP

即BOOPBEEP

 

4.[HDCTF2019]basic rsa

打开后：

 

 

 类型：n+e+c+p+q= m

直接套用脚本

import random
from binascii import a2b_hex,b2a_hex
p = 262248800182277040650192055439906580479
q = 262854994239322828547925595487519915551
n = p * q
def multiplicative_inversr(a,b):
    x = 0
    y = 1
    lx = 1
    ly = 0
    oa = a
    ob = b
    while b != 0:
        q = a // b
        (a, b) = (b, a % b)
        (x, lx) = ((lx - (q * x)), x)
        (y, ly) = ((ly - (q * y)), y)
    if lx < 0:
        lx += ob
    if ly < 0:
        ly += oa
    return lx
def gcd(a,b):
    while b != 0:
        a, b = b, a % b
    return a
def generate_keypair(p,q):
    n = p * q
    phi = (p - 1) * (q -1)
    e = 65533
    g = gcd(e, phi)
    while g != 1:
        e = random.randrange(1, phi)
        g = gcd(e, phi)
    d = multiplicative_inversr(e, phi)
    return ((e,n),(d,n))
def encrypt(pk, plaintext):
    key, n = pk[0]
    print(b2a_hex(plaintext.encode()))
    cipher = pow(int(b2a_hex(plaintext.encode()),16), key , n)
    return cipher
def decrypt(pk, cipher):
    key, n = pk[1]
    cipher = pow(cipher, key ,n)
    cipher = a2b_hex(hex(cipher).split('0x')[1])
    return cipher
pk = generate_keypair(p,q)
cipher = 27565231154623519221597938803435789010285480123476977081867877272451638645710
plaintext = decrypt(pk, cipher)
print(plaintext)

 

 

 

5.[GXYCTF2019]CheckIn

dikqTCpfRjA8fUBIMD5GNDkwMjNARkUwI0BFTg==

base64解密：

 

 

 再ROT47解码：

得到GXY{Y0u_kNow_much_about_Rot}

 

 

 

 

 

 

 

 

6.[GUET-CTF2019]BabyRSA

p+q : 0x1232fecb92adead91613e7d9ae5e36fe6bb765317d6ed38ad890b4073539a6231a6620584cea5730b5af83a3e80cf30141282c97be4400e33307573af6b25e2ea

(p+1)(q+1) : 0x5248becef1d925d45705a7302700d6a0ffe5877fddf9451a9c1181c4d82365806085fd86fbaab08b6fc66a967b2566d743c626547203b34ea3fdb1bc06dd3bb765fd8b919e3bd2cb15bc175c9498f9d9a0e216c2dde64d81255fa4c05a1ee619fc1fc505285a239e7bc655ec6605d9693078b800ee80931a7a0c84f33c851740

e : 0xe6b1bee47bd63f615c7d0a43c529d219

d : 0x2dde7fbaed477f6d62838d55b0d0964868cf6efb2c282a5f13e6008ce7317a24cb57aec49ef0d738919f47cdcd9677cd52ac2293ec5938aa198f962678b5cd0da344453f521a69b2ac03647cdd8339f4e38cec452d54e60698833d67f9315c02ddaa4c79ebaa902c605d7bda32ce970541b2d9a17d62b52df813b2fb0c5ab1a5

enc_flag : 0x50ae00623211ba6089ddfae21e204ab616f6c9d294e913550af3d66e85d0c0693ed53ed55c46d8cca1d7c2ad44839030df26b70f22a8567171a759b76fe5f07b3c5a6ec89117ed0a36c0950956b9cde880c575737f779143f921d745ac3bb0e379c05d9a3cc6bf0bea8aa91e4d5e752c7eb46b2e023edbc07d24a7c460a34a9a

import gmpy2
import libnum


a = 0x1232fecb92adead91613e7d9ae5e36fe6bb765317d6ed38ad890b4073539a6231a6620584cea5730b5af83a3e80cf30141282c97be4400e33307573af6b25e2ea
b = 0x5248becef1d925d45705a7302700d6a0ffe5877fddf9451a9c1181c4d82365806085fd86fbaab08b6fc66a967b2566d743c626547203b34ea3fdb1bc06dd3bb765fd8b919e3bd2cb15bc175c9498f9d9a0e216c2dde64d81255fa4c05a1ee619fc1fc505285a239e7bc655ec6605d9693078b800ee80931a7a0c84f33c851740
e = 0xe6b1bee47bd63f615c7d0a43c529d219
d = 0x2dde7fbaed477f6d62838d55b0d0964868cf6efb2c282a5f13e6008ce7317a24cb57aec49ef0d738919f47cdcd9677cd52ac2293ec5938aa198f962678b5cd0da344453f521a69b2ac03647cdd8339f4e38cec452d54e60698833d67f9315c02ddaa4c79ebaa902c605d7bda32ce970541b2d9a17d62b52df813b2fb0c5ab1a5
enc_flag = 0x50ae00623211ba6089ddfae21e204ab616f6c9d294e913550af3d66e85d0c0693ed53ed55c46d8cca1d7c2ad44839030df26b70f22a8567171a759b76fe5f07b3c5a6ec89117ed0a36c0950956b9cde880c575737f779143f921d745ac3bb0e379c05d9a3cc6bf0bea8aa91e4d5e752c7eb46b2e023edbc07d24a7c460a34a9a
n = b-a-1


M = pow(enc_flag, d, n)
print (libnum.n2s(M))

找到脚本

 

 flag{cc7490e-78ab-11e9-b422-8ba97e5da1fd}

 

7.二战时期，某国军官与一个音乐家情妇相好，然而自从那时起，他屡战屡败，敌人似乎料事如神。他也有怀疑过他的情妇，但是他经过24小时观察他的情妇，发现她每天都只是作曲，然后弹奏给战地电台，为士兵们鼓气，并未有任何逾越。那么，间谍到底是谁？这张曲谱是否有猫腻？ (答案为一个明文字符串，提交获得的有意义语句通顺字符串即可) 注意：得到的 flag 请包上 flag{} 提交

打开发现：

 

 

 图片中的线索很明显ASCLL码 八进制

数字三个一组，转换就行了

得到flag{ILoveSecurityVeryMuch}

假设e=3, e很小，但是n很大。
回顾RSA加密公式： C=M^e % n (C密文，M明文)