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 点击复制后，将打开C知道体验页

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。 ```c #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。

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： ```c #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。