帮我写一段无需使用相关库的带输出的N值较小的NTRU数字签名的C语言代码

好的，下面是一份简单的NTRU数字签名的C语言代码，不需要使用相关库：

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <time.h>

#define N 11
#define Q 64
#define P 3

int32_t f[N], g[N], h[N];

int32_t rand_coe();
int32_t mod_q(int64_t a);
int32_t mod_p(int64_t a);
int32_t poly_eval(int32_t *poly, int32_t x);
void poly_mult(int32_t *f, int32_t *g, int32_t *h);
void poly_add(int32_t *f, int32_t *g, int32_t *h);
void poly_sub(int32_t *f, int32_t *g, int32_t *h);
void poly_inv(int32_t *f, int32_t *g);

int main() {
    srand(time(NULL));
    int32_t p_inv = 0, e[n] = {0};
    int32_t r[n] = {0}, s[n] = {0}, m[n] = {0};
    int32_t m_hash[n] = {0}, f_hash[n] = {0}, g_hash[n] = {0};

    // Generate random f and g
    for (int i = 0; i < N; i++) {
        f[i] = rand_coe();
        g[i] = rand_coe();
    }
    // Compute h = f * g mod q
    poly_mult(f, g, h);

    // Randomly choose a message
    for (int i = 0; i < N; i++) {
        m[i] = rand_coe();
    }
    // Hash the message
    poly_mult(m, m, m_hash);
    poly_add(m_hash, h, m_hash);
    poly_mod_q(m_hash);
    // Compute f * g^-1 mod p
    poly_inv(g, g);
    poly_mult(f, g, f);
    poly_mod_p(f);
    // Compute (m + h * f * g^-1) mod p
    poly_mult(h, f, e);
    poly_add(e, m, e);
    poly_mod_p(e);

    // Generate random r
    for (int i = 0; i < N; i++) {
        r[i] = rand_coe();
    }
    // Compute s = r + e mod p
    poly_add(r, e, s);
    poly_mod_p(s);
    // Hash message, f, and g
    poly_mult(f, f, f_hash);
    poly_mult(g, g, g_hash);
    poly_add(m_hash, f_hash, m_hash);
    poly_add(m_hash, g_hash, m_hash);
    poly_mod_p(m_hash);
    // Compute p^-1 mod 3
    for (int i = 0; i < P-2; i++) {
        p_inv = mod_p(p_inv + P);
    }
    // Compute signature
    for (int i = 0; i < N; i++) {
        s[i] = mod_p(s[i] * p_inv);
    }

    // Print signature
    printf("Signature:\n");
    for (int i = 0; i < N; i++) {
        printf("%d ", s[i]);
    }
    printf("\n");

    return 0;
}

int32_t rand_coe() {
    return rand() % Q - Q / 2;
}

int32_t mod_q(int64_t a) {
    return ((a % Q) + Q) % Q;
}

int32_t mod_p(int64_t a) {
    return ((a % P) + P) % P;
}

int32_t poly_eval(int32_t *poly, int32_t x) {
    int32_t y = 0, x_pow = 1;
    for (int i = 0; i < N; i++) {
        y += poly[i] * x_pow;
        x_pow = mod_q(x_pow * x);
    }
    return y;
}

void poly_mult(int32_t *f, int32_t *g, int32_t *h) {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            h[(i+j)%N] += mod_q(f[i] * g[j]);
        }
    }
}

void poly_add(int32_t *f, int32_t *g, int32_t *h) {
    for (int i = 0; i < N; i++) {
        h[i] = mod_p(f[i] + g[i]);
    }
}

void poly_sub(int32_t *f, int32_t *g, int32_t *h) {
    for (int i = 0; i < N; i++) {
        h[i] = mod_p(f[i] - g[i]);
    }
}

void poly_inv(int32_t *f, int32_t *g) {
    int32_t v[N], r[N], s[N];
    for (int i = 0; i < N; i++) {
        v[i] = rand() % P;
        r[i] = f[i];
        s[i] = 0;
    }
    s[0] = 1;

    int32_t u, q, t[N];
    for (int i = 0; i < N; i++) {
        u = r[0];
        q = u ? mod_p(v[0] * r[0]) : 0;
        t[0] = s[N-1];
        for (int j = 1; j < N; j++) {
            t[j] = mod_p(s[j-1] - q * v[N-j]);
            r[j-1] = r[j];
            s[j-1] = s[j];
        }
        r[N-1] = u;
        s[N-1] = t[N-1];
        v[0] = v[N-1];
        for (int j = 1; j < N; j++) {
            v[j] = v[j-1];
        }
    }

    poly_mod_p(s);
    for (int i = 0; i < N; i++) {
        g[i] = s[i];
    }
}

该代码实现的是NTRU签名方案，其中N=11，Q=64，P=3。输入为一个长度为N的多项式m，输出为一个长度为N的多项式s，即签名。