帮我写一段无需使用相关库的带输出的NTRU数字签名的C语言代码

好的，以下是一段使用NTRU数字签名的C语言代码，不需要使用相关库。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <math.h>

#define N 509 // NTRU的参数
#define Q 2048
#define B 120
#define ITER 100 // NTRU签名的迭代次数

int Mod(int a, int b) {
    int ret = a % b;
    if (ret < 0)
        return ret + b;
    else
        return ret;
}

int gcd(int a, int b) {
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

int InvMod(int a, int b) {
    int b0 = b, t, q;
    int x0 = 0, x1 = 1;
    if (b == 1)
        return 1;
    while (a > 1) {
        q = a / b;
        t = b, b = a % b, a = t;
        t = x0, x0 = x1 - q * x0, x1 = t;
    }
    if (x1 < 0)
        x1 += b0;
    return x1;
}

int ModExp(int a, int b, int m) {
    int r = 1;
    while (b) {
        if (b & 1)
            r = (r * a) % m;
        a = (a * a) % m;
        b >>= 1;
    }
    return r;
}

int Rng(int bound) {
    return rand() % bound;
}

void ByteSwap(char* data, int len) {
    for (int i = 0; i < len / 2; i++) {
        char temp = data[i];
        data[i] = data[len - i - 1];
        data[len - i - 1] = temp;
    }
}

void Round(int* f, int q) {
    for (int i = 0; i < N; i++) {
        if (f[i] > q / 2)
            f[i] -= q;
        if (f[i] < -q / 2)
            f[i] += q;
    }
}

void Encrypt(int* ct, int* pt, int* key, int q) {
    int r[N], e[N];
    for (int i = 0; i < N; i++) {
        r[i] = Rng(2) - 1;
        e[i] = Rng(q / B) * B;
    }
    int b[N];
    for (int i = 0; i < N; i++) {
        b[i] = 0;
        for (int j = 0; j < N; j++)
            b[i] += key[i * N + j] * r[j];
        b[i] = Mod(b[i], q);
    }
    for (int i = 0; i < N; i++)
        ct[i] = Mod(pt[i] + b[i] + e[i], q);
}

int Verify(int* ct, int* sig, int* key, int q) {
    int pt[N], pt1[N];
    memset(pt1, 0, sizeof(pt1));
    Encrypt(pt, sig, key, q);
    int sum = 0;
    for (int i = 0; i < N; i++) {
        pt1[i] = Mod(ct[i] - pt[i], q);
        if (pt1[i] < -q / 4 || pt1[i] > q / 4)
            return 0;
        sum += abs(pt1[i]);
    }
    return (sum < B);
}

void Sign(int* sig, int* pt, int* key, int q) {
    int s[N], e[N], f[N], g[N];
    for (int i = 0; i < N; i++) {
        s[i] = Rng(2) - 1;
        e[i] = Rng(q / B) * B;
    }
    int a[N * N];
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (i == j)
                a[i * N + j] = q;
            else
                a[i * N + j] = 0;
        }
    }
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            a[i * N + j] += key[i * N + j];
        }
    }
    int x[N], y[N], z[N];
    for (int k = 0; k < N; k++) {
        x[k] = 0;
        for (int i = 0; i < N; i++)
            x[k] += a[k * N + i] * pt[i];
        x[k] = Mod(x[k], q);
        y[k] = 0;
        for (int i = 0; i < N; i++)
            y[k] += a[k * N + i] * s[i];
        y[k] = Mod(y[k], q);
        z[k] = Mod(x[k] - y[k] + e[k], q);
    }
    memcpy(f, sig, N * sizeof(int));
    for (int i = 0; i < ITER; i++) {
        for (int k = 0; k < N; k++) {
            g[k] = 0;
            for (int j = 0; j < N; j++) {
                g[k] += key[k * N + j] * f[j];
            }
            g[k] = Mod(g[k], q);
        }
        for (int k = 0; k < N; k++) {
            f[k] = Mod(f[k] - Mod(z[k] - g[k], q), q);
        }
        Round(f, q);
    }
    memcpy(sig, f, N * sizeof(int));
}

int main() {
    srand(time(NULL));
    int p = 3, q = 2048, d = 1, N = 509;
    int pt[N], sig[N], ct[N], key[N * N];
    memset(key, 0, sizeof(key));
    for (int i = 0; i < N; i++) {
        pt[i] = Rng(2) - 1;
    }
    int p1 = p + 1, invd = InvMod(d, q);
    while (1) {
        do {
            for (int i = 0; i < N; i++) {
                key[i * N + i] = Rng(2) - 1;
                for (int j = i + 1; j < N; j++) {
                    key[i * N + j] = Rng(p1) * invd;
                    key[j * N + i] = -key[i * N + j];
                }
            }
        } while (gcd(key[0], q) != 1);
        int flag = 1;
        for (int i = 0; i < ITER; i++) {
            Sign(sig, pt, key, q);
            if (!Verify(ct, sig, key, q)) {
                flag = 0;
                break;
            }
        }
        if (flag)
            break;
    }
    printf("pt: ");
    for (int i = 0; i < N; i++)
        printf("%d ", pt[i]);
    printf("\n");
    printf("sig: ");
    for (int i = 0; i < N; i++)
        printf("%d ", sig[i]);
    printf("\n");
    printf("ct: ");
    for (int i = 0; i < N; i++)
        printf("%d ", ct[i]);
    printf("\n");

    return 0;
}

该程序使用NTRU数字签名算法实现，在签名时先生成一个随机密钥，然后进行迭代计算得到签名，再使用该密钥对签名进行验证。程序中的参数已经设定为常用的NTRU参数，如果需要更改参数可以根据实际需求进行修改。