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#include <cstring> #include <iostream> #include <random> using int_t = long long int; using std::cin; using std::cout; using std::endl; const int_t LARGE = 2e6; const int_t mod = 998244353; const int_t g = 3; int_t power(int_t b, int_t i) { int_t r = 1; if (i < 0) i = (i % (mod - 1) + mod - 1) % (mod - 1); while (i) { if (i & 1) r = r * b % mod; b = b * b % mod; i >>= 1; } return r; } void makeflip(int_t* arr, int_t size2) { int_t len = (1 << size2); arr[0] = 0; for (int_t i = 1; i < len; i++) { arr[i] = (arr[i >> 1] >> 1) | ((i & 1) << (size2 - 1)); } } int_t upper2n(int_t x) { int_t r = 0; while ((1 << r) < x) r++; return r; } template <int_t arg = 1> void transform(int_t* A, int_t size2, int_t* flip) { int_t len = (1 << size2); for (int_t i = 0; i < len; i++) { int_t r = flip[i]; if (r > i) std::swap(A[i], A[r]); } for (int_t i = 2; i <= len; i *= 2) { int_t mr = power(g, arg * (mod - 1) / i); for (int_t j = 0; j < len; j += i) { int_t curr = 1; for (int_t k = 0; k < i / 2; k++) { int_t u = A[j + k], v = curr * A[j + k + i / 2] % mod; A[j + k] = (u + v) % mod; A[j + k + i / 2] = (u - v + mod) % mod; curr = curr * mr % mod; } } } } void poly_inv(const int_t* A, int_t n, int_t* result) { /* 这里的n是模x^n 计算B(x)*A(x) = 1 mod x^n, 其中A(x)已知 假设已知A(x)*B(x) = 1 mod x^{ceil(n/2)} 假设C(x)*A(x) = 1 mod x^n (A(x)B(x)-1)^2 = A^2(x)B^2(x)-2A(x)B(x)+1= 0 A(x)B^2(x)-2B(x)+C(x) = 0 C(x) = 2B(x)-A(x)B^2(x) */ // #ifdef DEBUG // cout << "at " << n << endl; // #endif if (n == 1) { result[0] = power(A[0], -1); return; } int_t next = n / 2 + n % 2; poly_inv(A, next, result); //次数不要选错了，应该用n次的A和B去卷 int_t size2 = upper2n(n + 2 * next + 1); static int_t X[LARGE]; static int_t Y[LARGE]; int_t len = (1 << size2); // 写错设置范围了！ memset(X + n, 0, sizeof(X[0]) * (len - n)); memset(Y + next, 0, sizeof(Y[0]) * (len - next)); memcpy(X, A, sizeof(A[0]) * n); memcpy(Y, result, sizeof(result[0]) * next); static int_t fliparr[LARGE]; makeflip(fliparr, size2); transform<1>(X, size2, fliparr); transform<1>(Y, size2, fliparr); for (int_t i = 0; i < len; i++) { X[i] = (2 * Y[i] - X[i] * Y[i] % mod * Y[i] % mod + mod) % mod; } transform<-1>(X, size2, fliparr); const int_t inv = power(len, -1); for (int_t i = 0; i < n; i++) result[i] = X[i] * inv % mod; #ifdef DEBUG cout << "poly inv at " << n << endl; cout << "result = "; for (int_t i = 0; i < next; i++) cout << result[i] << " "; cout << endl; #endif } int_t poly_divat(const int_t* F, const int_t* G, int_t n, int_t k) { /* n次多项式F和G 计算F(x)/G(x)的k次项前系数 考虑F(x)*G(-x)/G(x)*G(-x)，分母只有偶数次项，写为C(x^2)；分子写成xA(x^2)+B(x^2)，如果k是奇数，那么递归(A,C,n,k/2)，如果k是偶数，那么递归(B,C,n,k/2) 到k<=n时直接计算 */ int_t size2 = upper2n(2 * n + 1); int_t len = 1 << size2; static int_t fliparr[LARGE]; makeflip(fliparr, size2); static int_t T1[LARGE], T2[LARGE], T3[LARGE]; for (int_t i = 0; i < len; i++) { if (i <= n) T1[i] = F[i], T2[i] = G[i]; else T1[i] = T2[i] = T3[i] = 0; } const int_t inv = power(len, -1); while (k >= n) { #ifdef DEBUG cout << "curr at k = " << k << endl; #endif for (int_t i = 0; i < len; i++) { if (i <= n) { T3[i] = T2[i] * (i % 2 ? (mod - 1) : 1); } else T3[i] = 0; } transform(T1, size2, fliparr); transform(T2, size2, fliparr); transform(T3, size2, fliparr); for (int_t i = 0; i < len; i++) { T1[i] = T1[i] * T3[i] % mod; T2[i] = T2[i] * T3[i] % mod; } transform<-1>(T1, size2, fliparr); transform<-1>(T2, size2, fliparr); #ifdef DEBUG cout << "prod T1 = "; for (int_t i = 0; i < len; i++) cout << T1[i] * inv % mod << " "; cout << endl; cout << "prod T2 = "; for (int_t i = 0; i < len; i++) cout << T2[i] * inv % mod << " "; cout << endl; #endif for (int_t i = 0; i < len; i++) { if (i * 2 < len) { T2[i] = T2[i * 2] * inv % mod; } else T2[i] = 0; } int_t b = k % 2; for (int_t i = 0; i < len; i++) { if (i % 2 == b) { T1[i / 2] = T1[i] * inv % mod; #ifdef DEBUG cout << "at " << i << " assgin " << T1[i] * inv << " to " << i / 2 << endl; #endif } #ifdef DEBUG cout << "assign 0 to " << i << endl; #endif if (i > 0) T1[i] = 0; } #ifdef DEBUG cout << "finished T1 = "; for (int_t i = 0; i < len; i++) cout << T1[i] % mod << " "; cout << endl; cout << "finished T2 = "; for (int_t i = 0; i < len; i++) cout << T2[i] % mod << " "; cout << endl; #endif k >>= 1; } poly_inv(T2, k + 1, T3); // #ifdef DEBUG // cout << "finished k = " << k << endl; // cout << "T1 = "; // for (int_t i = 0; i < len; i++) // cout << T1[i] % mod << " "; // cout << endl; // cout << "T2 = "; // for (int_t i = 0; i < len; i++) // cout << T2[i] % mod << " "; // cout << endl; // cout << "T2 inv = "; // for (int_t i = 0; i < len; i++) // cout << T3[i] % mod << " "; // cout << endl; // #endif int_t result = 0; //计算结果的k次项 for (int_t i = 0; i <= k; i++) { result = (result + T1[i] * T3[k - i] % mod) % mod; } return result; } void poly_mul(const int_t* A, int_t n, const int_t* B, int_t m, int_t* C) { /* 计算n次多项式A与m次多项式B的乘积 */ int_t size2 = upper2n(n + m + 1); int_t len = 1 << size2; static int_t T1[LARGE], T2[LARGE]; for (int_t i = 0; i < len; i++) { if (i <= n) T1[i] = A[i]; else T1[i] = 0; if (i <= m) T2[i] = B[i]; else T2[i] = 0; } static int_t fliparr[LARGE]; makeflip(fliparr, size2); transform(T1, size2, fliparr); transform(T2, size2, fliparr); for (int_t i = 0; i < len; i++) T1[i] = T1[i] * T2[i] % mod; transform<-1>(T1, size2, fliparr); int_t inv = power(len, -1); for (int_t i = 0; i <= n + m; i++) C[i] = T1[i] * inv % mod; } int_t poly_linear_rec(const int_t* A0, const int_t* F0, int_t n, int_t k) { /* 计算线性递推 F[1],F[2]...F[k] 递推系数 A[0],A[1]...A[k-1] 首项 A[m]=A[m-1]*F[1]+A[m-2]*F[2]+...+A[m-k]*F[k] */ static int_t T1[LARGE], T2[LARGE]; static int_t Ax[LARGE], Fx[LARGE]; Fx[0] = 0; Ax[k] = 0; for (int i = 1; i <= k; i++) { Fx[i] = F0[i]; Ax[i - 1] = A0[i - 1]; } poly_mul(Ax, k, Fx, k, T1); T1[0] = Ax[0]; for (int i = 1; i <= k - 1; i++) { T1[i] = (Ax[i] - T1[i] + mod) % mod; } for (int i = k; i <= 2 * k; i++) T1[i] = 0; T2[0] = 1; for (int i = 1; i <= k; i++) T2[i] = (mod - Fx[i]) % mod; // #ifdef DEBUG // cout << "T1 = "; // for (int_t i = 0; i <= k; i++) { // cout << T1[i] << " "; // } // cout << endl; // cout << "T2 = "; // for (int_t i = 0; i <= k; i++) { // cout << T2[i] << " "; // } // cout << endl; // #endif return poly_divat(T1, T2, k, n); } void poly_log(const int_t* A, int_t n, int_t* out) { /* 计算log(A(x)), A(x)为n次多项式 DlogF(x) =DF(x)/F(x) */ static int_t Ad[LARGE]; static int_t Finv[LARGE]; static int_t R[LARGE]; const int_t m = n - 1; for (int_t i = 0; i <= m; i++) { Ad[i] = A[i + 1] * (i + 1) % mod; } Ad[n] = 0; poly_inv(A, n + 1, Finv); poly_mul(Ad, m, Finv, n, R); for (int_t i = 1; i <= n; i++) { out[i] = R[i - 1] * power(i, -1) % mod; } } void poly_exp(const int_t* A, int_t n, int_t* out) { /* 计算exp(A(x)), A(x)为n次多项式 H(x)=G(x)(1-logG(x)+A(x)) H(x)为当次递推项，G(x)为上一次递推项 */ if (n == 1) { out[0] = 1; return; } int_t r = n / 2 + n % 2; poly_exp(A, r, out); static int_t G[LARGE]; static int_t G2[LARGE]; static int_t logG[LARGE]; static int_t R[LARGE]; for (int_t i = 0; i < r; i++) G[i] = out[i]; for (int_t i = r; i < n; i++) G[i] = 0; poly_log(G, n - 1, logG); // for (int_t i = r; i < n; i++) // logG[i] = 0; for (int_t i = 0; i < n; i++) { G2[i] = (mod - logG[i] + A[i]) % mod; } G2[0] = (G2[0] + 1) % mod; poly_mul(G, n - 1, G2, n - 1, R); for (int_t i = 0; i < n; i++) out[i] = R[i]; #ifdef DEBUG cout << "at " << n << endl; cout << "A = "; for (int_t i = 0; i < n; i++) cout << A[i] << " "; cout << endl; cout << "G = "; for (int_t i = 0; i < n; i++) cout << G[i] << " "; cout << endl; cout << "logG = "; for (int_t i = 0; i < n; i++) cout << logG[i] << " "; cout << endl; cout << "G2 = "; for (int_t i = 0; i < n; i++) cout << G2[i] << " "; cout << endl; cout << "R = "; for (int_t i = 0; i < n; i++) cout << R[i] << " "; cout << endl; cout << endl; #endif } int main() { std::ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); int_t n; cin >> n; static int_t A[LARGE], B[LARGE]; static int_t C[LARGE]; for (int_t i = 0; i < n; i++) cin >> A[i]; poly_exp(A, n, B); for (int_t i = 0; i < n; i++) cout << B[i] << " "; return 0; } |