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NSSRound16

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NSSCTF Round16 Crypto题目

Pr

题目

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from Crypto.Util.number import *
import random

flag=plaintext = 'NSSCTF{****************}'
charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
padding_length = 100 - len(plaintext)

for _ in range(padding_length):
    plaintext += random.choice(charset)

public_exponent = 31413537523
message = bytes_to_long(plaintext.encode())
assert message > (1 << 512)
assert message < (1 << 1024)

prime_p = getPrime(512)
prime_q = getPrime(512)
prime_r = getPrime(512)
n1 = prime_p * prime_q
n2 = prime_q * prime_r
ciphertext1 = pow(message, public_exponent, n1)
ciphertext2 = pow(message, public_exponent, n2)
print('c1=', ciphertext1)
print('c2=', ciphertext2)
print('p=', prime_p)
print('r=', prime_r)
'''
c1= 36918910341116680090654563538246204134840776220077189276689868322808977412566781872132517635399441578464309667998925236488280867210758507758915311644529399878185776345227817559234605958783077866016808605942558810445187434690812992072238407431218047312484354859724174751718700409405142819140636116559320641695
c2= 15601788304485903964195122196382181273808496834343051747331984997977255326224514191280515875796224074672957848566506948553165091090701291545031857563686815297483181025074113978465751897596411324331847008870832527695258040104858667684793196948970048750296571273364559767074262996595282324974180754813257013752
p= 12101696894052331138951718202838643670037274599483776996203693662637821825873973767235442427190607145999472731101517998719984942030184683388441121181962123
r= 10199001137987151966640837133782537428248507382360655526592866939552984259171772190788036403425837649697437126360866173688083643144865107648483668545682383
'''

题解

根据题目可知，明文的范围在512-1024之间 转化式子然后利用中国剩余定理就能得到 之后，就在模pr的情况下解c就行

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from Crypto.Util.number import *  # 一个非常好用的crypto库
import gmpy2

public_exponent = 31413537523
c1 = 36918910341116680090654563538246204134840776220077189276689868322808977412566781872132517635399441578464309667998925236488280867210758507758915311644529399878185776345227817559234605958783077866016808605942558810445187434690812992072238407431218047312484354859724174751718700409405142819140636116559320641695
c2 = 15601788304485903964195122196382181273808496834343051747331984997977255326224514191280515875796224074672957848566506948553165091090701291545031857563686815297483181025074113978465751897596411324331847008870832527695258040104858667684793196948970048750296571273364559767074262996595282324974180754813257013752
p = 12101696894052331138951718202838643670037274599483776996203693662637821825873973767235442427190607145999472731101517998719984942030184683388441121181962123
r = 10199001137987151966640837133782537428248507382360655526592866939552984259171772190788036403425837649697437126360866173688083643144865107648483668545682383


# b是密文列表  m是模数列表
def crt(b, m):
    # 传入的参数分别为密文 和 模数n
    # 乘积计算
    M = 1
    for i in range(len(m)):
        M *= m[i]

    Mm = []
    # 求余数 M/m[i]
    for i in range(len(m)):
        Mm.append(M // m[i])

    # 求Mm[i]的乘法逆元
    Mm_ = []
    for i in range(len(m)):
        t, a, _ = gmpy2.gcdext(Mm[i], m[i])
        Mm_.append(int(a % m[i]))

    # 求的累加
    y = 0
    for i in range(len(m)):
        y += (Mm[i] * Mm_[i] * b[i])
    y = y % M
    return y, M


c3, n = crt([c1, c2], [p, r])
print(c3)
print(n)
phi = (p - 1) * (r - 1)
d = gmpy2.invert(public_exponent, phi)
m = pow(c3, d, n)
print(long_to_bytes(m))

break

题目

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Bc8tSTrvGJm2oYuCzIz+Yg4nwwKBgQDiYUawe5Y+rPbFhVOMVB8ZByfMa4LjeSDd
Z23jEGvylBHSeyvFCQq3ISUE40k1D2XmmeaZML3a1nUn6ORIWGaG2phcwrWLkR6n
ubVmb1QJSzgzmFHGnL56KHByZxD9q6DPB+o6gGWt8/6ddBl2NIZU/1btdPQgojfA
XXJFzR92RQKBgQC7qlB0U7m2U4FdG9eelSd+WSKNUVllZAuHji7jgh7Ox6La9xN5
miGZ1yvP44yX218OJ9Zi08o6vIrM6Eil45KzTtGm4iuIn8CMpox+5eUtoxyvxa9r
s2Wu+IRZN9zCME+p+qI8/TG27dIyDzsdgNqcUo8ESls7uW5/FEA7bYTCiQKBgQC7
1KybeB+kZ0zlfIdi8tVOpeI+uaHDbdh3+/5wHUsD3hmfg7VAag0q/2RA1vkB/oG1
QVLVHl0Yu0I/1/u5jyeakrtClAegAsvlrK+3i321rGS4YpTPb3SX1P/f3GZ7o7Ds
touA+NHk8IL9T7xkmJYw5h/RLG32ucH6aU6MXfLR5QKBgD/skfdFxGWxhHk6U1mS
27IM9jJNg9xLz5nxzkqPPhLn+rdgIIuTuQtv++eEjEP++7ZV10rg5yKVJd/bxy8H
2IN7aQo7kZWulHTQDZMFwgOhn0u6glJi+qC8bWzYDFOQSFrY9XQ3vwKMspqm+697
xM+dMUW0LML6oUE9ZjEiAY/5
-----END PRIVATE KEY-----

题解

第一个文件是一个不完整的私钥证书，第二个就是密文

对私钥先base64解码，然后转化为16进制进行关键数据恢复，分割符是0x0281

第一个数据就是q，第二个数据是dp，第三个数据是dq，第四个数据是p在q上的逆元

这里没有给出n，但是q的位数已经是1024位了，可以尝试在模q上求解 则

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from Crypto.Util.number import *

c = 6081370370545409218106271903400346695565292992689150366474451604281551878507114813906275593034729563149286993189430514737137534129570304832172520820901940874698337733991868650159489601159238582002010625666203730677577976307606665760650563172302688129824842780090723167480409842707790983962415315804311334507726664838464859751689906850572044873633896253285381878416855505301919877714965930289139921111644393144686543207867970807469735534838601255712764863973853116693691206791007433101433703535127367245739289103650669095061417223994665200039533840922696282929063608853551346533188464573323230476645532002621795338655
q = 0x00e26146b07b963eacf6c585538c541f190727cc6b82e37920dd676de3106bf29411d27b2bc5090ab7212504e349350f65e699e69930bddad67527e8e448586686da985cc2b58b911ea7b9b5666f54094b38339851c69cbe7a2870726710fdaba0cf07ea3a8065adf3fe9d741976348654ff56ed74f420a237c05d7245cd1f7645
dp = 0x00bbaa507453b9b653815d1bd79e95277e59228d515965640b878e2ee3821ecec7a2daf713799a2199d72bcfe38c97db5f0e27d662d3ca3abc8acce848a5e392b34ed1a6e22b889fc08ca68c7ee5e52da31cafc5af6bb365aef8845937dcc2304fa9faa23cfd31b6edd2320f3b1d80da9c528f044a5b3bb96e7f14403b6d84c289
dq = 0x00bbd4ac9b781fa4674ce57c8762f2d54ea5e23eb9a1c36dd877fbfe701d4b03de199f83b5406a0d2aff6440d6f901fe81b54152d51e5d18bb423fd7fbb98f279a92bb429407a002cbe5acafb78b7db5ac64b86294cf6f7497d4ffdfdc667ba3b0ecb68b80f8d1e4f082fd4fbc64989630e61fd12c6df6b9c1fa694e8c5df2d1e5
print(q.bit_length())
m = pow(c, dq, q)
print(long_to_bytes(m))