题目分析

这道题上来先把加密脚本用单表替换密码加密了，解开后又分为共模攻击和p为n1,n2公因数的两种类型RSA，解出密文后有一个对明文顺序的加密。

解题

频率分析

分析出来的脚本有个别地方不准确，手动更改一下

from gmpy2 import is_prime 
from os import urandom 
import base64 
def bytes_to_num(b): 
    return int(b.encode('hex'), 16) 
def num_to_bytes(n): 
    b = hex(n)[2:-1] 
    b = '0' + b 
    if len(b)%2 == 1 
        else b 
    return b.decode('hex') 
def get_a_prime(l): 
    random_seed = urandom(l) 
    num = bytes_to_num(random_seed) 
    while True: 
        if is_prime(num): 
            break 
        num+=1 
    return num 
def encrypt(s, e, n): 
    p = bytes_to_num(s) 
    p = pow(p, e, n) 
    return num_to_bytes(p).encode('hex') 
def separate(n): 
    p = n % 4 
    t = (p*p) % 4 
    return t == 1 
f = open('flag.txt', 'r') 
flag = f.read() 
msg1 = "" 
msg2 = "" 
for i in range(len(flag)): 
    if separate(i): 
        msg2 += flag[i] 
    else: msg1 += flag[i] 
p1 = get_a_prime(128) 
p2 = get_a_prime(128) 
p3 = get_a_prime(128) 
n1 = p1*p2 
n2 = p1*p3 
e = 0x1001 
c1 = encrypt(msg1, e, n1) 
c2 = encrypt(msg2, e, n2) 
print(c1) 
print(c2) 
e1 = 0x1001 
e2 = 0x101 
p4 = get_a_prime(128) 
p5 = get_a_prime(128) 
n3 = p4*p5 
c1 = num_to_bytes(pow(n1, e1, n3)).encode('hex') 
c2 = num_to_bytes(pow(n1, e2, n3)).encode('hex') 
print(c1) 
print(c2) 
print(base64.b64encode(num_to_bytes(n2))) 
print(base64.b64encode(num_to_bytes(n3)))

解题

①从题中可以看出,n1和n2的公因数为p1，我们只要知道p1,p2,p3这三个数，就可以求解得到msg1和msg2,，题中所给的文件里并没有给出n1,解出n1成为了解题的关键。

p1=gcd(n1,n2)
p2=n1//p1
p3=n2//p1
e = 0x1001
d1=invert(e,(p1-1)*(p2-1))
d2=invert(e,(p1-1)*(p3-1))
msg1=pow(c3,d1,n1)
msg2=pow(c4,d2,n2)

②向后看可以发现，后边关于n1的加密是共模攻击，按照共模攻击的套路解出n1

e1 = 0x1001 
e2 = 0x101 
s1,s2=exgcdmoni.exgcd(e1,e2)
n1=powmod(c1,s1,n3)*powmod(c2,s2,n3)%n3

③解出msg1和msg2,把加密函数逆回去，这里需要m1和m2两个计数器，分别控制msg1和msg2

msg1=(long_to_bytes(msg1)).decode()
msg2=(long_to_bytes(msg2)).decode()
def separate(n): 
    p = n % 4 
    t = (p*p) % 4 
    return t == 1
flag=''
m1=0
m2=0
for i in range(0,len(msg1)+len(msg2)):
    if separate(i):
        flag+=msg2[m1]
        m1+=1
    else:
        flag+=msg1[m2]
        m2+=1
print(flag)

完整代码

from gmpy2 import *
from Crypto.Util.number import *
import base64,exgcdmoni
c1=0x2639c28e3609a4a8c953cca9c326e8e062756305ae8aee6efcd346458aade3ee8c2106ab9dfe5f470804f366af738aa493fd2dc26cb249a922e121287f3eddec0ed8dea89747dc57aed7cd2089d75c23a69bf601f490a64f73f6a583081ae3a7ed52238c13a95d3322065adba9053ee5b12f1de1873dbad9fbf4a50a2f58088df0fddfe2ed8ca1118c81268c8c0fd5572494276f4e48b5eb424f116e6f5e9d66da1b6b3a8f102539b690c1636e82906a46f3c5434d5b04ed7938861f8d453908970eccef07bf13f723d6fdd26a61be8b9462d0ddfbedc91886df194ea022e56c1780aa6c76b9f1c7d5ea743dc75cec3c805324e90ea577fa396a1effdafa3090
c2=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
c3=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
c4=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
n2='PVNHb2BfGAnmxLrbKhgsYXRwWIL9eOj6K0s3I0slKHCTXTAUtZh3T0r+RoSlhpO3+77AY8P7WETYz2Jzuv5FV/mMODoFrM5fMyQsNt90VynR6J3Jv+fnPJPsm2hJ1Fqt7EKaVRwCbt6a4BdcRoHJsYN/+eh7k/X+FL5XM7viyvQxyFawQrhSV79FIoX6xfjtGW+uAeVF7DScRcl49dlwODhFD7SeLqzoYDJPIQS+VSb3YtvrDgdV+EhuS1bfWvkkXRijlJEpLrgWYmMdfsYX8u/+Ylf5xcBGn3hv1YhQrBCg77AHuUF2w/gJ/ADHFiMcH3ux3nqOsuwnbGSr7jA6Cw=='
n3='TmNVbWUhCXR1od3gBpM+HGMKK/4ErfIKITxomQ/QmNCZlzmmsNyPXQBiMEeUB8udO7lWjQTYGjD6k21xjThHTNDG4z6C2cNNPz73VIaNTGz0hrh6CmqDowFbyrk+rv53QSkVKPa8EZnFKwGz9B3zXimm1D+01cov7V/ZDfrHrEjsDkgK4ZlrQxPpZAPl+yqGlRK8soBKhY/PF3/GjbquRYeYKbagpUmWOhLnF4/+DP33ve/EpaSAPirZXzf8hyatL4/5tAZ0uNq9W6T4GoMG+N7aS2GeyUA2sLJMHymW4cFK5l5kUvjslRdXOHTmz5eHxqIV6TmSBQRgovUijlNamQ=='
n2=bytes_to_long(base64.b64decode(n2))
n3=bytes_to_long(base64.b64decode(n3))
e1 = 0x1001 
e2 = 0x101 
s1,s2=exgcdmoni.exgcd(e1,e2)
n1=powmod(c1,s1,n3)*powmod(c2,s2,n3)%n3
p1=gcd(n1,n2)
p2=n1//p1
p3=n2//p1
e = 0x1001
d1=invert(e,(p1-1)*(p2-1))
d2=invert(e,(p1-1)*(p3-1))
msg1=pow(c3,d1,n1)
msg2=pow(c4,d2,n2)
msg1=(long_to_bytes(msg1)).decode()
msg2=(long_to_bytes(msg2)).decode()
def separate(n): 
    p = n % 4 
    t = (p*p) % 4 
    return t == 1
flag=''
m1=0
m2=0
for i in range(0,len(msg1)+len(msg2)):
    if separate(i):
        flag+=msg2[m1]
        m1+=1
    else:
        flag+=msg1[m2]
        m2+=1
print(flag)

遇到的问题

这里发现字节类型的直接对其索引，会返回Int型数值

b=b'adkfh'
print(b[0])
>>97

所以最后一步解密时记得转成字符串