帮我写一段带输出的NTRU数字签名的python代码

好的，下面是一段带输出的NTRU数字签名的python代码：

import random

# NTRU参数
N = 503
p = 3
q = 64

# NTT变换
def ntt(f):
    w = 1
    for i in range(9):
        for j in range(1<<i):
            t = w*f[j|(1<<i)]
            f[j|(1<<i)] = f[j]-t
            f[j] += t
        w = (w*p) % q

# NTT逆变换
def intt(f):
    ntt(f)
    inv_n = pow(1<<9, q-2, q)
    for i in range(1<<9):
        f[i] = (f[i]*inv_n) % q
    for i in range(256):
        f[i] = (f[i>>1] >> (i&1) * 8) % 256

# NTRU密钥生成
def ntru_keygen():
    f = [random.randint(0, 1) for _ in range(N)]
    g = [random.randint(0, 1) for _ in range(N)]
    f[0] = 1
    g[0] = 1
    h = [(x-y)%q for x, y in zip(ntt(f), ntt(g))]
    intt(h)
    return f, g, h

# NTRU签名
def ntru_sign(m, f, g, h):
    r = [random.randint(0, 1) for _ in range(N)]
    e = [(x-y)%q for x, y in zip(ntt(r), ntt(h))]
    intt(e)
    f_prime = [(x*y)%q for x, y in zip(f, r)]
    e_prime = [(x*y)%q for x, y in zip(e, g)]
    c = [(x+y)%q for x, y in zip(ntt(f_prime), ntt(e_prime))]
    m_hash = hash(m) % (1<<8)
    s = [(m_hash-x)%q for x in ntt(c)]
    return s, c

# NTRU数字签名验证
def ntru_verify(m, s, c, g):
    m_hash = hash(m) % (1<<8)
    t = [(m_hash-x)%q for x in ntt(c)]
    return s == t

# 测试
m = 'hello world'
f, g, h = ntru_keygen()
s, c = ntru_sign(m, f, g, h)
print('Digital signature s:', s)
print('Public signature c:', c)
if ntru_verify(m, s, c, g):
    print('Correct signature!')
else:
    print('Wrong signature!')

注意，以上代码仅供参考，实际使用时需要根据项目需要进行修改和完善。