RSA加密OAEP方案

guoxiaomi · 发表于 2018-4-19 16:56:52

加入我们，或者，欢迎回来。

您需要登录才可以下载或查看，没有帐号？注册会员

x

本帖最后由 guoxiaomi 于 2018-4-27 23:56 编辑

RSA加密算法是一种非对称的加密手段，由于只是用来加密对称加密的秘钥（对于AES加密来说，128位的密钥足以保证安全），只会在传输密钥的时候执行一次，实际的运算量也不是很大，而且 ruby 自带 Bignum，写起来也方便些

1楼是数学原理，实用的生成秘钥对、加密方案在3、4楼，在RM中使用的话，需要用dll弄一个计算sha-1的函数（见5楼）。
2楼是C#的代码
6楼是针对 256-bit RSA密钥的攻击，仅仅花了5分钟，说明取更大的密钥的必要性

如此就可以用完成~~大概安全的~~公钥加密传输了。

下面是正文：

首先，从网上抄了3个函数，用来求模反和幂模

RUBY 代码复制

def egcd(a, b)
    if a == 0
        [b, 0, 1]
    else
        g, y, x = egcd(b % a, a)
        [g, x - (b / a) * y, y]
    end
end
 
def modinv(a, m)
    g, x, y = egcd(a, m)
    if g == 1
        x % m
    else
        raise
    end
end
 
def modpow(a, b, n)
    rec = 1
    while (b != 0)
        if b & 1 != 0
            rec = (rec * a) % n
        end
        a = (a * a) % n
        b = b >> 1
    end
    rec % n
end

def egcd(a, b)


    if a == 0


        [b, 0, 1]


    else


        g, y, x = egcd(b % a, a)


        [g, x - (b / a) * y, y]


    end


end


 


def modinv(a, m)


    g, x, y = egcd(a, m)


    if g == 1


        x % m


    else


        raise


    end


end


 


def modpow(a, b, n)


    rec = 1


    while (b != 0)


        if b & 1 != 0


            rec = (rec * a) % n


        end


        a = (a * a) % n


        b = b >> 1


    end


    rec % n


end

然后登陆 www.wolframalpha.com，用类似 prime closest to 2^256 * 1.23456789 的指令得到几个很大的质数，比如下面的几个例子

RUBY 代码复制

# p = 340282366920938463463374607431768211507 # 128-bit
# q = 442367076997220017613959734844163358521 # 128-bit
# p = 114321529704002274578921084903793614162317645586699719492822694265655811637217 # 256-bit
# q = 170095105326940384212383024501862619327831639589164392574626096589661701930901 # 256-bit
p = 39391335229695556645551048713636649037690892144144766626387899639258047233085306918331113608222769414886203283497951067975814999508178451226732581907071099 # 512-bit
q = 8091612085720357837842476099645383223098769278647227328308631524827867677878737958899289093390100825439815397081936915739099782156677184746940143587294947 # 512-bit
 
n = p * q
e = 65537
 
str = '1234567890abcdef'
m = str.unpack('C*').inject(0){|s, i| (s << 8) + i}
p m
p c = modpow(m, e, n) # public key: e, n
 
d = modinv(e, (p-1) * (q-1)) # private key: d, n
p m2 = modpow(c, d, n), m == m2

# p = 340282366920938463463374607431768211507 # 128-bit


# q = 442367076997220017613959734844163358521 # 128-bit


# p = 114321529704002274578921084903793614162317645586699719492822694265655811637217 # 256-bit


# q = 170095105326940384212383024501862619327831639589164392574626096589661701930901 # 256-bit


p = 39391335229695556645551048713636649037690892144144766626387899639258047233085306918331113608222769414886203283497951067975814999508178451226732581907071099 # 512-bit


q = 8091612085720357837842476099645383223098769278647227328308631524827867677878737958899289093390100825439815397081936915739099782156677184746940143587294947 # 512-bit


 


n = p * q


e = 65537


 


str = '1234567890abcdef'


m = str.unpack('C*').inject(0){|s, i| (s << 8) + i}


p m


p c = modpow(m, e, n) # public key: e, n


 


d = modinv(e, (p-1) * (q-1)) # private key: d, n


p m2 = modpow(c, d, n), m == m2

提醒一下，m是明文，c是密文，用m生成c只需要知道 e = 65537 和 n = p * q
解密的话需要知道 (p-1)*(q-1) 和 n，也就是 p 和 q，而用 n 求 p、q 几乎是不可能的
正常的加密要求 m < n，这里 n ~ 1024 bit 是足够长的
上面的内容显示结果：

65392825175610104412735852254614283622
55038560310780067437062090146057515347452012654059325469405078165508476134024808261020076587516103483241454346122071285542898770131400418286780269941688027757584931240339204470841864636279890281715398177884324364339943921202813369696583386135438024241244203849337673874382470838957284345740774426383985682694
65392825175610104412735852254614283622
true

复制代码

qweytr_1 · 发表于 2018-4-27 22:38:34

guoxiaomi 发表于 2018-4-27 12:02
（强行互动）用于计算sha-1的dll在我写的SAE服务器里就弄好了：hash.dll，输入一个字符串返回40位的hex码， ...

刚刚试了一下，指望128位质数的RSA还不如不加密……
建议至少按照标准用2048位的两个质数组合一个4096位的公钥
用的是例子里面的两个质数，然后把公钥送进msieve
In[1]:= p = 340282366920938463463374607431768211507
Out[1]= 340282366920938463463374607431768211507
In[2]:= q = 442367076997220017613959734844163358521
Out[2]= 442367076997220017613959734844163358521
In[3]:= p q
Out[3]= 150529716008511059192844022140063161898374205466644527991444838913354598701147
执行
msieve.exe -t 6 150529716008511059192844022140063161898374205466644527991444838913354598701147
得到msieve.log

Fri Apr 27 22:32:24 2018
Fri Apr 27 22:32:24 2018
Fri Apr 27 22:32:24 2018 Msieve v. 1.46
Fri Apr 27 22:32:24 2018 random seeds: 24c07960 fef15915
Fri Apr 27 22:32:24 2018 factoring 150529716008511059192844022140063161898374205466644527991444838913354598701147 (78 digits)
Fri Apr 27 22:32:24 2018 no P-1/P+1/ECM available, skipping
Fri Apr 27 22:32:24 2018 commencing quadratic sieve (78-digit input)
Fri Apr 27 22:32:24 2018 using multiplier of 3
Fri Apr 27 22:32:24 2018 using 32kb Intel Core sieve core
Fri Apr 27 22:32:24 2018 sieve interval: 12 blocks of size 32768
Fri Apr 27 22:32:24 2018 processing polynomials in batches of 17
Fri Apr 27 22:32:24 2018 using a sieve bound of 959759 (37824 primes)
Fri Apr 27 22:32:24 2018 using large prime bound of 95975900 (26 bits)
Fri Apr 27 22:32:24 2018 using trial factoring cutoff of 27 bits
Fri Apr 27 22:32:24 2018 polynomial 'A' values have 10 factors
Fri Apr 27 22:37:00 2018 37988 relations (19473 full + 18515 combined from 206527 partial), need 37920
Fri Apr 27 22:37:00 2018 begin with 226000 relations
Fri Apr 27 22:37:00 2018 reduce to 54259 relations in 2 passes
Fri Apr 27 22:37:00 2018 attempting to read 54259 relations
Fri Apr 27 22:37:00 2018 recovered 54259 relations
Fri Apr 27 22:37:00 2018 recovered 44076 polynomials
Fri Apr 27 22:37:00 2018 attempting to build 37988 cycles
Fri Apr 27 22:37:00 2018 found 37988 cycles in 1 passes
Fri Apr 27 22:37:00 2018 distribution of cycle lengths:
Fri Apr 27 22:37:00 2018 length 1 : 19473
Fri Apr 27 22:37:00 2018 length 2 : 18515
Fri Apr 27 22:37:00 2018 largest cycle: 2 relations
Fri Apr 27 22:37:01 2018 matrix is 37824 x 37988 (5.5 MB) with weight 1137187 (29.94/col)
Fri Apr 27 22:37:01 2018 sparse part has weight 1137187 (29.94/col)
Fri Apr 27 22:37:01 2018 filtering completed in 3 passes
Fri Apr 27 22:37:01 2018 matrix is 27054 x 27118 (4.3 MB) with weight 900229 (33.20/col)
Fri Apr 27 22:37:01 2018 sparse part has weight 900229 (33.20/col)
Fri Apr 27 22:37:01 2018 saving the first 48 matrix rows for later
Fri Apr 27 22:37:01 2018 matrix is 27006 x 27118 (2.8 MB) with weight 652841 (24.07/col)
Fri Apr 27 22:37:01 2018 sparse part has weight 452517 (16.69/col)
Fri Apr 27 22:37:01 2018 matrix includes 64 packed rows
Fri Apr 27 22:37:01 2018 commencing Lanczos iteration
Fri Apr 27 22:37:01 2018 memory use: 4.0 MB
Fri Apr 27 22:37:08 2018 lanczos halted after 429 iterations (dim = 27006)
Fri Apr 27 22:37:08 2018 recovered 18 nontrivial dependencies
Fri Apr 27 22:37:08 2018 prp39 factor: 340282366920938463463374607431768211507
Fri Apr 27 22:37:08 2018 prp39 factor: 442367076997220017613959734844163358521
Fri Apr 27 22:37:08 2018 elapsed time 00:04:44

复制代码

可以看出，破解256bit公钥（2个128bit质数相乘只用五分钟）

guoxiaomi · 发表于 2018-4-27 12:02:17

（强行互动）用于计算sha-1的dll在我写的SAE服务器里就弄好了：hash.dll，输入一个字符串返回40位的hex码，所以用的时候要转换一下~

RUBY 代码复制

SHA1 = Win32API.new('hash.dll', 'sha1_v', ['p', 'p', 'i'], 'v')
def sha1(s)
  buffer = "\0" * 40
  SHA1.call(buffer, string, string.size)
  [buffer].pack('H*')
end

SHA1 = Win32API.new('hash.dll', 'sha1_v', ['p', 'p', 'i'], 'v')


def sha1(s)


  buffer = "\0" * 40


  SHA1.call(buffer, string, string.size)


  [buffer].pack('H*')


end

guoxiaomi · 发表于 2018-4-27 11:43:40

本帖最后由 guoxiaomi 于 2018-4-27 11:45 编辑

RUBY 代码复制

# ---------------------------------------------------------
# RSA
# ---------------------------------------------------------
# For OAEP:
# [url]https://www.emc.com/collateral/white-papers/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp.pdf[/url]
# ---------------------------------------------------------
 
require 'base64'
require 'digest/sha1'
 
module RSA
  def int2str(i)
    hex = i.to_s(16)
    hex = '0' + hex if hex.size % 2 == 1
    [hex].pack('H*')
  end
 
  def str2int(s)
    s.unpack('H*')[0].to_i(16)
  end
 
  def load_pub(fn)
    s = File.open(fn, 'r') do |f|
      Base64.decode64 f.read.split(/\s+/)[1]
    end
 
    i = 0
    vars = []
    while i < s.length do
      length = str2int(s[i...i+4])
      i += 4
      vars.push s[i...i+length]
      i += length
    end
 
    if i == s.size
      e, n = str2int(vars[1]), str2int(vars[2])
      return [e, n]
    else
      return false
    end
  end
 
  def egcd(a, b)
    if a == 0
        [b, 0, 1]
    else
        g, y, x = egcd(b % a, a)
        [g, x - (b / a) * y, y]
    end
  end
 
  def modinv(a, m)
      g, x, y = egcd(a, m)
      if g == 1
          x % m
      else
          raise
      end
  end
 
  def modpow(a, b, n)
      rec = 1
      while (b != 0)
          if b & 1 != 0
              rec = (rec * a) % n
          end
          a = (a * a) % n
          b = b >> 1
      end
      rec % n
  end
 
  def i2osp(i, len)
    a = []
    len.times do
      a.push(i & 255)
      i = i >> 8
    end
    a.reverse.pack('C*')
  end
 
  alias os2ip str2int
 
  def xor(s1, s2)
    raise if s1.size != s2.size
    int2str(str2int(s1) ^ str2int(s2))
  end
 
  def sha1(s)
    Digest::SHA1.digest(s)
  end
 
  def mgf1(s, l)
    t = ''
    for i in 0..(l / 20) # 20 is sha1 output length
      t << sha1(s + i2osp(i, 4))
    end
    t[0...l]
  end
 
  alias encrypt_public modpow
end
 
include RSA
 
# try to encrypt a 128-bit aes key:
aes_key = 'Say:hello,world!'
# ---------------------------------------------------------
# 0. load public key
# ---------------------------------------------------------
e, n = load_pub('test.pub')
 
# ---------------------------------------------------------
# 1. length checking
# ---------------------------------------------------------
L = ''
DB = ''
k = int2str(n).size
lHash = sha1(L)
hLen = lHash.size
M = aes_key
mLen = M.size
if hLen > 2 << 61 - 1 # check for sha-1
  raise 'label too long'
elsif mLen + 2 * hLen + 2 > k
  raise 'message too long'
end
 
# ---------------------------------------------------------
# 2. EME-OAEP encoding
# ---------------------------------------------------------
PS = "\x00" * (k - 2 * hLen - mLen - 2)
DB << (lHash + PS + "\x01" + M)
seed = Array.new(hLen).collect{|e| rand(256)}.pack('C*')
maskedDB = xor(DB, mgf1(seed, k - hLen - 1))
maskedseed = xor(seed, mgf1(maskedDB, hLen))
 
EM = "\x00" + maskedseed + maskedDB
print Base64.encode64(EM)
 
# ---------------------------------------------------------
# 3. RSA encryption
# ---------------------------------------------------------
m = os2ip(EM)
c = encrypt_public(m, e, n)
C = i2osp(c, k)
print Base64.encode64(C)

# ---------------------------------------------------------


# RSA


# ---------------------------------------------------------


# For OAEP:


# [url]https://www.emc.com/collateral/white-papers/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp.pdf[/url]


# ---------------------------------------------------------


 


require 'base64'


require 'digest/sha1'


 


module RSA


  def int2str(i)


    hex = i.to_s(16)


    hex = '0' + hex if hex.size % 2 == 1


    [hex].pack('H*')


  end


 


  def str2int(s)


    s.unpack('H*')[0].to_i(16)


  end


 


  def load_pub(fn)


    s = File.open(fn, 'r') do |f|


      Base64.decode64 f.read.split(/\s+/)[1]


    end


 


    i = 0


    vars = []


    while i < s.length do


      length = str2int(s[i...i+4])


      i += 4


      vars.push s[i...i+length]


      i += length


    end


 


    if i == s.size


      e, n = str2int(vars[1]), str2int(vars[2])


      return [e, n]


    else


      return false


    end


  end


 


  def egcd(a, b)


    if a == 0


        [b, 0, 1]


    else


        g, y, x = egcd(b % a, a)


        [g, x - (b / a) * y, y]


    end


  end


 


  def modinv(a, m)


      g, x, y = egcd(a, m)


      if g == 1


          x % m


      else


          raise


      end


  end


 


  def modpow(a, b, n)


      rec = 1


      while (b != 0)


          if b & 1 != 0


              rec = (rec * a) % n


          end


          a = (a * a) % n


          b = b >> 1


      end


      rec % n


  end


 


  def i2osp(i, len)


    a = []


    len.times do


      a.push(i & 255)


      i = i >> 8


    end


    a.reverse.pack('C*')


  end


 


  alias os2ip str2int


 


  def xor(s1, s2)


    raise if s1.size != s2.size


    int2str(str2int(s1) ^ str2int(s2))


  end


 


  def sha1(s)


    Digest::SHA1.digest(s)


  end


 


  def mgf1(s, l)


    t = ''


    for i in 0..(l / 20) # 20 is sha1 output length


      t << sha1(s + i2osp(i, 4))


    end


    t[0...l]


  end


 


  alias encrypt_public modpow


end


 


include RSA


 


# try to encrypt a 128-bit aes key:


aes_key = 'Say:hello,world!'


# ---------------------------------------------------------


# 0. load public key


# ---------------------------------------------------------


e, n = load_pub('test.pub')


 


# ---------------------------------------------------------


# 1. length checking


# ---------------------------------------------------------


L = ''


DB = ''


k = int2str(n).size


lHash = sha1(L)


hLen = lHash.size


M = aes_key


mLen = M.size


if hLen > 2 << 61 - 1 # check for sha-1


  raise 'label too long'


elsif mLen + 2 * hLen + 2 > k


  raise 'message too long'


end


 


# ---------------------------------------------------------


# 2. EME-OAEP encoding


# ---------------------------------------------------------


PS = "\x00" * (k - 2 * hLen - mLen - 2)


DB << (lHash + PS + "\x01" + M)


seed = Array.new(hLen).collect{|e| rand(256)}.pack('C*')


maskedDB = xor(DB, mgf1(seed, k - hLen - 1))


maskedseed = xor(seed, mgf1(maskedDB, hLen))


 


EM = "\x00" + maskedseed + maskedDB


print Base64.encode64(EM)


 


# ---------------------------------------------------------


# 3. RSA encryption


# ---------------------------------------------------------


m = os2ip(EM)


c = encrypt_public(m, e, n)


C = i2osp(c, k)


print Base64.encode64(C)

PHP的测试代码，把ruby代码的输出 Base64.encode64(C) 复制到 $data里，直接在服务端把私钥写出来了：

PHP 代码复制

<?php
    $key = "-----BEGIN RSA PRIVATE KEY-----
MIIEpAIBAAKCAQEAw2pCSxFVb2g1JwKjXK7CT7cVuSqo5emnWKyhjeqQjSvDUS3x
dn1k4uTq24c701kpHkdXYl2WCJlDAJrIYD1KH6dzFAEOXNWKlKWIkcF7YkvnhZnn
eHJw5ZhYuTM7cligkMeFW8UWb8tI/SsBmCQZrA/spJ03JmqJtoIY9PQXO+XAGqCy
ZLxnHz40bvRMncgjPm4XENojrVtzh9LdRei8N3RkZw4gEcVaPpPgJze2ObSKdZ4+
CUtuFOSdap7K7o2N2d8S/Y7Jd+hH84FnSMpKlwsrr5unWeSFE67iNJV/KDH7V7Su
5J1GB9WgOBkxOm86+EXhrQfeuBI0Jqh/HQPKlwIDAQABAoIBAHkQqcWMFZwRuWxT
rBXNh0CPsvumvEDEQxMi0Gv1uFxdkoIGrqO0B/6xx/b9j0je4iSKGk8sn7jwCpyj
SCSnk/aMbTUlelp1SYfSDgwPDHvXfrY8pTTv4ZW8jqseXGP5GHDZy3sRXVchYDTI
ohI5u1Xn/7n0VKiONIdHjmKjnolSNj8DRGYJ/nL9u4QenX0tGN43/lqP7UhO/O7D
OdU9ysLn7glad6EARmMbG8hxIavdLvJ6F5HS9tpVT+9jTgi9t4mqdP6YTeh10fWa
KQ7HFCOb9Wmu84YRNgb1+9w+ARGoe6r+aO1HTtE2DtoycExWOL0Ic2zhtKpKdv7N
8eZc7iECgYEA5Kuou03+IKlwDizgWAvsCifSWxRFcUAoCvR/xnDF80wqSxVWtJTF
KaPHqU6uNTlFHvda+KRy0N6g8bc0+nrF8ZaOzMMM7tZTtY29mp5dZawD7zODTEGa
XXzDbLyA8+rW3XZpuHlXJH6rovFyXvcZK+w5ekmeJI0NMwxTlvkUUK8CgYEA2sUf
6wZ2A/+FMjaqNWM5Nii/vth9F3VIQgpNC45batRyr2xSJBrug6Jd9R/N5qizWJR8
7/KYHwalqbqufj74PwodGRCR8QWoIK6msOEnIPNzhr+iFvFesjMvfF2MmPYzBT40
U+aJjqYS39rQRJuKlg6GcVEYiGu3J0Dc0LVqDpkCgYABGkLKBGpBRg/gQZ8jTYtp
0R9WiRZ8oU6QHvWuw0RxE+DwU74DSORaewuvaU21u/z1VUP/Buv2zdwAzl1XB9iE
fbFak2YwkJ/+tBxB7pmMr/Ok556uc3KHGN7oW1BT3MIEd1mzJgKhjmrNclSW2KIq
cA0m5cv5aSSzJlAQ3kqWyQKBgQCOL96931Ym5RVJ/JOF3Xaax8NQI96xt21+Vrma
kCzEc04SzAFbLBHnhkkw/znQ98aRBPlq7q6GQ8i6VvYAYabxOf6NguKH11hd3YXD
oqLO3MwyQSz1Ym1cvr3XQ+oUpLemabaS7VxsVW4hBlOks79QJiKlVcLvL6s7nQGn
uLE0yQKBgQDAtCjdKbGGo8OFg2FKFD5vGXSXMscchi0ImyN4gAxRK3rlx5fyoo+X
WUFSHdC3vO9TM00S07eXIuaOaHWjQUOzoTXKmQk4bvf5nZy6u+NvuPwGmwXXCCD2
Z+tK5j2eJ6DkOCwms9IA5kISZkxcFQASlywTihpkmXW9V9mXB0vk5g==
-----END RSA PRIVATE KEY-----
";
    $data = "gEg4eOsosLi3LxguBSsyz7GVM/TBZo8XZYH97YYdNes9oL/1RYbbYp7jnfVt
IWQpaeXFl/PBg+d1vy2R0C1YSuvqKJ/h5a6SddgNCF0sY40T3Pwg1n6bHcjp
AXy+lrduQqnAUivppJF0UTY6Bv6DvdzQ5odd5On1tcZPzX/oGTAEYMiCZRWm
OCqDPa7CqO2ohDaY0v+Ke8pxZh/my/VeyxVjprvkV3CWm4fE0DTwAnvSJtSJ
6zJV0iEMsigfdSk9CnWPkCkeaHHCbj9mQ25u+bjtoFAw+FpQAzB9CF6YQHI/
hCe4BPr8odOxdZtyKhgtdyENOmdPRWewtIhPc7PtJw==";
    echo $data;
    $data = base64_decode($data);
    openssl_private_decrypt($data, $decrypted, $key, OPENSSL_NO_PADDING);
    echo "<br>NOPADDING!<br>";
    echo base64_encode($decrypted);
    echo "<br>";
    openssl_private_decrypt($data, $decrypted2, $key, OPENSSL_PKCS1_OAEP_PADDING);
    echo "<br>OAEP!<br>";
    echo $decrypted2."<br>";
    echo base64_encode($decrypted2);
?>

<?php


    $key = "-----BEGIN RSA PRIVATE KEY-----


MIIEpAIBAAKCAQEAw2pCSxFVb2g1JwKjXK7CT7cVuSqo5emnWKyhjeqQjSvDUS3x


dn1k4uTq24c701kpHkdXYl2WCJlDAJrIYD1KH6dzFAEOXNWKlKWIkcF7YkvnhZnn


eHJw5ZhYuTM7cligkMeFW8UWb8tI/SsBmCQZrA/spJ03JmqJtoIY9PQXO+XAGqCy


ZLxnHz40bvRMncgjPm4XENojrVtzh9LdRei8N3RkZw4gEcVaPpPgJze2ObSKdZ4+


CUtuFOSdap7K7o2N2d8S/Y7Jd+hH84FnSMpKlwsrr5unWeSFE67iNJV/KDH7V7Su


5J1GB9WgOBkxOm86+EXhrQfeuBI0Jqh/HQPKlwIDAQABAoIBAHkQqcWMFZwRuWxT


rBXNh0CPsvumvEDEQxMi0Gv1uFxdkoIGrqO0B/6xx/b9j0je4iSKGk8sn7jwCpyj


SCSnk/aMbTUlelp1SYfSDgwPDHvXfrY8pTTv4ZW8jqseXGP5GHDZy3sRXVchYDTI


ohI5u1Xn/7n0VKiONIdHjmKjnolSNj8DRGYJ/nL9u4QenX0tGN43/lqP7UhO/O7D


OdU9ysLn7glad6EARmMbG8hxIavdLvJ6F5HS9tpVT+9jTgi9t4mqdP6YTeh10fWa


KQ7HFCOb9Wmu84YRNgb1+9w+ARGoe6r+aO1HTtE2DtoycExWOL0Ic2zhtKpKdv7N


8eZc7iECgYEA5Kuou03+IKlwDizgWAvsCifSWxRFcUAoCvR/xnDF80wqSxVWtJTF


KaPHqU6uNTlFHvda+KRy0N6g8bc0+nrF8ZaOzMMM7tZTtY29mp5dZawD7zODTEGa


XXzDbLyA8+rW3XZpuHlXJH6rovFyXvcZK+w5ekmeJI0NMwxTlvkUUK8CgYEA2sUf


6wZ2A/+FMjaqNWM5Nii/vth9F3VIQgpNC45batRyr2xSJBrug6Jd9R/N5qizWJR8


7/KYHwalqbqufj74PwodGRCR8QWoIK6msOEnIPNzhr+iFvFesjMvfF2MmPYzBT40


U+aJjqYS39rQRJuKlg6GcVEYiGu3J0Dc0LVqDpkCgYABGkLKBGpBRg/gQZ8jTYtp


0R9WiRZ8oU6QHvWuw0RxE+DwU74DSORaewuvaU21u/z1VUP/Buv2zdwAzl1XB9iE


fbFak2YwkJ/+tBxB7pmMr/Ok556uc3KHGN7oW1BT3MIEd1mzJgKhjmrNclSW2KIq


cA0m5cv5aSSzJlAQ3kqWyQKBgQCOL96931Ym5RVJ/JOF3Xaax8NQI96xt21+Vrma


kCzEc04SzAFbLBHnhkkw/znQ98aRBPlq7q6GQ8i6VvYAYabxOf6NguKH11hd3YXD


oqLO3MwyQSz1Ym1cvr3XQ+oUpLemabaS7VxsVW4hBlOks79QJiKlVcLvL6s7nQGn


uLE0yQKBgQDAtCjdKbGGo8OFg2FKFD5vGXSXMscchi0ImyN4gAxRK3rlx5fyoo+X


WUFSHdC3vO9TM00S07eXIuaOaHWjQUOzoTXKmQk4bvf5nZy6u+NvuPwGmwXXCCD2


Z+tK5j2eJ6DkOCwms9IA5kISZkxcFQASlywTihpkmXW9V9mXB0vk5g==


-----END RSA PRIVATE KEY-----


";


    $data = "gEg4eOsosLi3LxguBSsyz7GVM/TBZo8XZYH97YYdNes9oL/1RYbbYp7jnfVt


IWQpaeXFl/PBg+d1vy2R0C1YSuvqKJ/h5a6SddgNCF0sY40T3Pwg1n6bHcjp


AXy+lrduQqnAUivppJF0UTY6Bv6DvdzQ5odd5On1tcZPzX/oGTAEYMiCZRWm


OCqDPa7CqO2ohDaY0v+Ke8pxZh/my/VeyxVjprvkV3CWm4fE0DTwAnvSJtSJ


6zJV0iEMsigfdSk9CnWPkCkeaHHCbj9mQ25u+bjtoFAw+FpQAzB9CF6YQHI/


hCe4BPr8odOxdZtyKhgtdyENOmdPRWewtIhPc7PtJw==";


    echo $data;


    $data = base64_decode($data);


    openssl_private_decrypt($data, $decrypted, $key, OPENSSL_NO_PADDING);


    echo "<br>NOPADDING!<br>";


    echo base64_encode($decrypted);


    echo "<br>";


    openssl_private_decrypt($data, $decrypted2, $key, OPENSSL_PKCS1_OAEP_PADDING);


    echo "<br>OAEP!<br>";


    echo $decrypted2."<br>";


    echo base64_encode($decrypted2);


?>

公钥：

TXT 代码复制

ssh-rsa AAAAB3NzaC1yc2EAAAADAQABAAABAQDDakJLEVVvaDUnAqNcrsJPtxW5Kqjl6adYrKGN6pCNK8NRLfF2fWTi5OrbhzvTWSkeR1diXZYImUMAmshgPUofp3MUAQ5c1YqUpYiRwXtiS+eFmed4cnDlmFi5MztyWKCQx4VbxRZvy0j9KwGYJBmsD+yknTcmaom2ghj09Bc75cAaoLJkvGcfPjRu9EydyCM+bhcQ2iOtW3OH0t1F6Lw3dGRnDiARxVo+k+AnN7Y5tIp1nj4JS24U5J1qnsrujY3Z3xL9jsl36EfzgWdIykqXCyuvm6dZ5IUTruI0lX8oMftXtK7knUYH1aA4GTE6bzr4ReGtB964EjQmqH8dA8qX test

ssh-rsa AAAAB3NzaC1yc2EAAAADAQABAAABAQDDakJLEVVvaDUnAqNcrsJPtxW5Kqjl6adYrKGN6pCNK8NRLfF2fWTi5OrbhzvTWSkeR1diXZYImUMAmshgPUofp3MUAQ5c1YqUpYiRwXtiS+eFmed4cnDlmFi5MztyWKCQx4VbxRZvy0j9KwGYJBmsD+yknTcmaom2ghj09Bc75cAaoLJkvGcfPjRu9EydyCM+bhcQ2iOtW3OH0t1F6Lw3dGRnDiARxVo+k+AnN7Y5tIp1nj4JS24U5J1qnsrujY3Z3xL9jsl36EfzgWdIykqXCyuvm6dZ5IUTruI0lX8oMftXtK7knUYH1aA4GTE6bzr4ReGtB964EjQmqH8dA8qX test

guoxiaomi · 发表于 2018-4-23 11:39:43

利用 ssh-keygen 生成 rsa 的密钥：

ssh-keygen -t rsa -f ./test -C 'test'

复制代码

用下面的脚本读取公钥（test.pub）中的e和n，在rpg maker中的不需要 require 'base64' ：

RUBY 代码复制

require 'base64'
 
s = File.open('test.pub', 'r') do |f|
    Base64.decode64 f.read.split(/\s+/)[1]
end
 
i = 0
vars = []
while i < s.length do
    length = s[i...i+4].unpack('H*')[0].to_i(16)
    i += 4
    vars.push s[i...i+length]
    i += length
end
p vars[0], i == s.length
p e = vars[1].unpack('H*')[0].to_i(16)
p n = vars[2].unpack('H*')[0].to_i(16)

require 'base64'


 


s = File.open('test.pub', 'r') do |f|


    Base64.decode64 f.read.split(/\s+/)[1]


end


 


i = 0


vars = []


while i < s.length do


    length = s[i...i+4].unpack('H*')[0].to_i(16)


    i += 4


    vars.push s[i...i+length]


    i += length


end


p vars[0], i == s.length


p e = vars[1].unpack('H*')[0].to_i(16)


p n = vars[2].unpack('H*')[0].to_i(16)

如此就可以得到 e 和 n
解析私钥（test）用：

openssl rsa -in test -text

复制代码

反正保存私钥的服务端肯定有办法去做解密的~

不死鸟之翼 · 发表于 2018-4-22 21:25:42

本帖最后由不死鸟之翼于 2018-4-22 21:35 编辑

冒个泡顺便C#大法好啥都有）

CSHARP 代码复制

namespace RSATest
{
    class Program
    {
        static RSAParameters GenKeyPair()
        {
            var RSA = new RSACryptoServiceProvider();
            return RSA.ExportParameters(true);
        }
 
        static byte[] Encrypt(byte[] m, byte[] n, byte[] e)
        {
            var RSA = new RSACryptoServiceProvider();
            var param = new RSAParameters()
            {
                Modulus = n,
                Exponent = e
            };
            RSA.ImportParameters(param);
            return RSA.Encrypt(m, false);
        }
 
        static byte[] Decrypt(byte[] c, RSAParameters param)
        {
            var RSA = new RSACryptoServiceProvider();
            RSA.ImportParameters(param);
            return RSA.Decrypt(c, false);
        }
 
        static void Main(string[] args)
        {
            var param = GenKeyPair();
            var s = "1234567890abcdef";
            var m = Encoding.UTF8.GetBytes(s);
            var c = Encrypt(m, param.Modulus, param.Exponent);
            var m2 = Decrypt(c, param);
            var s2 = Encoding.UTF8.GetString(m2);
            Console.WriteLine(s);
        }
    }
}

namespace RSATest


{


    class Program


    {


        static RSAParameters GenKeyPair()


        {


            var RSA = new RSACryptoServiceProvider();


            return RSA.ExportParameters(true);


        }


 


        static byte[] Encrypt(byte[] m, byte[] n, byte[] e)


        {


            var RSA = new RSACryptoServiceProvider();


            var param = new RSAParameters()


            {


                Modulus = n,


                Exponent = e


            };


            RSA.ImportParameters(param);


            return RSA.Encrypt(m, false);


        }


 


        static byte[] Decrypt(byte[] c, RSAParameters param)


        {


            var RSA = new RSACryptoServiceProvider();


            RSA.ImportParameters(param);


            return RSA.Decrypt(c, false);


        }


 


        static void Main(string[] args)


        {


            var param = GenKeyPair();


            var s = "1234567890abcdef";


            var m = Encoding.UTF8.GetBytes(s);


            var c = Encrypt(m, param.Modulus, param.Exponent);


            var m2 = Decrypt(c, param);


            var s2 = Encoding.UTF8.GetString(m2);


            Console.WriteLine(s);


        }


    }


}

一楼所述的textbook RSA（没有填充）用来展示数学原理没问题，但实际使用会有安全性问题，业界实践都是有padding的比如OAEP（最优非对称加密填充）
所以用C#你会发现每次加密的结果都不一样）

另外没C#的话还可以用Windows Powershell 反正都是.net）雾

POWERSHELL 代码复制

$rsa=New-Object -TypeName "System.Security.Cryptography.RSACryptoServiceProvider"

$arr=[System.Text.Encoding]::UTF8.GetBytes("1234567890abcdef")

echo $rsa.Encrypt($arr,$false)

$rsa=New-Object -TypeName "System.Security.Cryptography.RSACryptoServiceProvider"



$arr=[System.Text.Encoding]::UTF8.GetBytes("1234567890abcdef")



echo $rsa.Encrypt($arr,$false)

帐号		自动登录	找回密码
密码			注册会员

[原创发布] RSA加密OAEP方案

加入我们，或者，欢迎回来。

评分

点评

评分

点评

点评

点评

评分

开拓者