用 Python 来实现 RSA 加解密

p = genprime(1024)
q = genprime(1024)

n = p * q
lambda_n = (p - 1) * (q - 1)

e = 35537

d = eucalg(e, lambda_n)[0]
if d < 0: d += lambda_n
return (d, n), (e, n)

def create_keys():
 p = genprime(1024)
 q = genprime(1024)
 n = p * q
 lambda_n = (p - 1) * (q - 1)
 e = 35537
 d = eucalg(e, lambda_n)[0]
 if d < 0: d += lambda_n
 return (d, n), (e, n)

def encrypt_data(data,key):
e_data = []
for d in data:
 e = modpow(d, key[0], key[1]) 
 e_data.append(e)
return e_data

## 加密和解密的逻辑完全一样
decrypt_data = encrypt_data

def modpow(b, e, n):
 # find length of e in bits
 tst = 1
 siz = 0
 while e >= tst:
tst <<= 1
siz += 1
 siz -= 1
 # calculate the result
 r = 1
 for i in range(siz, -1, -1):
r = (r * r) % n
if (e >> i) & 1: r = (r * b) % n
 return r

# matrix multiplication
def sqmatrixmul(m1, m2, w, mod):
 mr = [[0 for j in range(w)] for i in range(w)]
 for i in range(w):
for j in range(w):
 for k in range(w):
mr[i][j] = (mr[i][j] + m1[i][k] * m2[k][j]) % mod
 return mr

# fibonacci calculator
def fib(x, mod):
 if x < 3: return 1
 x -= 2
 # find length of e in bits
 tst = 1
 siz = 0
 while x >= tst:
tst <<= 1
siz += 1
 siz -= 1
 # calculate the matrix
 fm = [
# function matrix
[0, 1],
[1, 1]
 ]
 rm = [
# result matrix
# (identity)
[1, 0],
[0, 1]
 ]
 for i in range(siz, -1, -1):
rm = sqmatrixmul(rm, rm, 2, mod)
if (x >> i) & 1:
 rm = sqmatrixmul(rm, fm, 2, mod)

 # second row of resulting vector is result
 return (rm[1][0] + rm[1][1]) % mod

def genprime(siz):
 while True:
num = (1 << (siz - 1)) + secrets.randbits(siz - 1) - 10;
# num must be 3 or 7 (mod 10)
num -= num % 10
num += 3 # 3 (mod 10)
# heuristic test
if modpow(2, num - 1, num) == 1 and fib(num + 1, num) == 0:
 return num
num += 5 # 7 (mod 10)
# heuristic test
if modpow(2, num - 1, num) == 1 and fib(num + 1, num) == 0:
 return num

e * x - lambda_n * y =1

def eucalg(a, b):
 # make a the bigger one and b the lesser one
 swapped = False
 if a < b:
a, b = b, a
swapped = True
 # ca and cb store current a and b in fORM of
 # coefficients with initial a and b
 # a' = ca[0] * a + ca[1] * b
 # b' = cb[0] * a + cb[1] * b
 ca = (1, 0)
 cb = (0, 1)
 while b != 0:
# k denotes how many times number b
# can be substracted from a
k = a // b
# swap a and b so b is always the lesser one
a, b, ca, cb = b, a-b*k, cb, (ca[0]-k*cb[0], ca[1]-k*cb[1])
 if swapped:
return (ca[1], ca[0])
 else:
return ca

python test.py make-keys rsakey

python test.py encrypt 明文.txt from rsakey to 密文.txt

python test.py decrypt 密文.txt as rsakey to 解密后.txt