python实现sm2加解密算法，不使用库

实现SM2算法需要用到椭圆曲线加密算法和哈希算法，这里我们使用Python内置的`hashlib`和`cryptomath`模块来实现。

首先，我们需要定义椭圆曲线的参数和基点：

P = 0xFFFFFFFEFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00000000FFFFFFFFFFFFFFFF
A = 0xFFFFFFFEFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00000000FFFFFFFFFFFFFFFC
B = 0x28E9FA9E9D9F5E344D5A9E4BCF6509A7F39789F515AB8F92DDBCBD414D940E93
N = 0xFFFFFFFEFFFFFFFFFFFFFFFFFFFFFFFF7203DF6B21C6052B53BBF40939D54123
Gx = 0x32C4AE2C1F1981195F9904466A39C9948FE30BBFF2660BE171BEEB6E3A9D4F5B
Gy = 0xBC3736A2F4F6779C59BDCEE36B692153D0A9877CC62A474002DF32E52139F0A

然后，我们需要实现点加和点倍运算：

def add(p1, p2):
    if p1 is None:
        return p2
    if p2 is None:
        return p1
    x1, y1 = p1
    x2, y2 = p2
    if x1 == x2 and y1 != y2:
        return None
    if x1 == x2:
        m = (3 * x1 * x1 + A) * cryptomath.invmod(2 * y1, P) % P
    else:
        m = (y1 - y2) * cryptomath.invmod(x1 - x2, P) % P
    x3 = (m * m - x1 - x2) % P
    y3 = (m * (x1 - x3) - y1) % P
    return (x3, y3)

def mul(k, p):
    if k == 0 or p is None:
        return None
    if k == 1:
        return p
    if k % 2 == 0:
        return mul(k // 2, add(p, p))
    else:
        return add(p, mul(k - 1, p))

接下来，我们需要实现签名和验签的函数：

def sign(msg, d):
    e = int(hashlib.sha256(msg).hexdigest(), 16)
    k = func.random_int_range(1, N)
    while True:
        x, y = mul(k, (Gx, Gy))
        r = (e + x) % N
        if r == 0 or k >= N:
            k = func.random_int_range(1, N)
            continue
        s = (cryptomath.invmod(k, N) * (d * r + k * e)) % N
        if s != 0:
            break
        k = func.random_int_range(1, N)
    return r, s

def verify(msg, q, r, s):
    e = int(hashlib.sha256(msg).hexdigest(), 16)
    if r <= 0 or r >= N or s <= 0 or s >= N:
        return False
    t = (r + s) % N
    if t == 0:
        return False
    x, y = add(mul(s, (Gx, Gy)), mul(t, q))
    if x is None or y is None:
        return False
    if (r + x) % N == e:
        return True
    else:
        return False

最后，我们可以用以下代码来测试SM2算法的加解密：

d = func.random_int_range(1, N)
q = mul(d, (Gx, Gy))
msg = b'Hello, world!'

# 签名和验签
r, s = sign(msg, d)
if verify(msg, q, r, s):
    print('Signature verified.')
else:
    print('Signature verification failed.')

# 加密和解密
plaintext = b'Hello, world!'
k = func.random_int_range(1, N)
x1, y1 = mul(k, (Gx, Gy))
c1 = hex(x1)[2:].rjust(64, '0') + hex(y1)[2:].rjust(64, '0')
s = mul(N - d, add((int(c1[:64], 16), int(c1[64:], 16)), mul(k, plaintext)))
c2 = hex(s[0])[2:].rjust(64, '0') + hex(s[1])[2:].rjust(64, '0')
ciphertext = bytes.fromhex(c1 + c2)
x2, y2 = mul(d, (int(ciphertext[:64], 16), int(ciphertext[64:128], 16)))
decrypted_text = (s[1] - y2) // x2
print(decrypted_text.to_bytes((decrypted_text.bit_length() + 7) // 8, 'big'))

需要注意的是，这里生成的私钥和公钥都是随机生成的，加密和解密的明文和密文都是字节串。