感谢绿哥
/* author: purinliang */ const int MAXLOGN = 20; const int MAXN = 1 << MAXLOGN; const int MOD = 998244353; const int G = 3; inline int qadd(const int &x, const int &y) { int r = x + y; return r >= MOD ? r - MOD : r; } inline int qsub(const int &x, const int &y) { int r = x - y; return r < 0 ? r + MOD : r; } inline int qmul(const int &x, const int &y) { ll r = 1LL * x * y; if(r >= MOD) r %= MOD; return r; } int qpow(int x, int n) { ll res = 1; while(n) { if(n & 1) res = qmul(res, x); x = qmul(x, x); n >>= 1; } return res; } using poly_t = int[MAXN]; using poly = int *const; /* Show h[0...n-1], be careful that the length of h is less than n. */ void polyshow(poly &h, int n) { for(int i = 0; i != n; ++i) printf("%d%c", h[i], " "[i == n - 1]); } void FNTT(poly &h, int n, int op) { static int gl[MAXLOGN], rev[MAXN], revl; if(gl[0] == 0) { for(int l = 0; l < MAXLOGN; ++l) gl[l] = qpow(G, (MOD - 1) / (1 << (l + 1))); } if((1 << revl) != n) { for(revl = 0; (1 << revl) != n; ++revl); for(int i = 0; i < n; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (revl - 1)); } for(int i = 0; i < n; ++i) { if(i < rev[i]) swap(h[i], h[rev[i]]); } for(int l = 0, len; (len = 1 << l) < n; ++l) { for(int i = 0, Gl = gl[l]; i < n; i += (len << 1)) { for(int j = i, w = 1; j < i + len; ++j, w = qmul(w, Gl)) { int u = h[j], t = qmul(h[j + len], w); h[j] = qadd(u, t), h[j + len] = qsub(u, t); } } } if(op == -1) { reverse(h + 1, h + n); for(int i = 0, inv = qpow(n, MOD - 2); i < n; ++i) h[i] = qmul(h[i], inv); } } /* Enlarge n to the smallest power of 2. */ void pretreat(poly &h, int &n) { int tn = 1; while(tn < n) tn <<= 1; fill(h + n, h + tn, 0); n = tn; } /* Add h1 and h2, and store the result in f. */ int polyadd(poly &h1, int n1, poly &h2, int n2, poly &f) { int n = max(n1, n2); fill(h1 + n1, h1 + n, 0); fill(h2 + n2, h2 + n, 0); for(int i = 0; i != n; ++i) f[i] = qadd(h1[i], h2[i]); return n; } /* Substract h2 from h1, and store the result in f. */ int polysub(poly &h1, int n1, poly &h2, int n2, poly &f) { int n = max(n1, n2); fill(h1 + n1, h1 + n, 0); fill(h2 + n2, h2 + n, 0); for(int i = 0; i != n; ++i) f[i] = qsub(h1[i], h2[i]); return n; } /* Multiply h1 and h2, and store the result in f. */ int polymul(poly &h1, int n1, poly &h2, int n2, poly &f) { int n = n1 + n2 - 1, tn = 1; while(tn < n) tn <<= 1; fill(h1 + n1, h1 + tn, 0), FNTT(h1, tn, 1); fill(h2 + n2, h2 + tn, 0), FNTT(h2, tn, 1); for(int i = 0; i != tn; ++i) f[i] = qmul(h1[i], h2[i]); FNTT(f, tn, -1); return n; } /* The following methods are solved in the sense of modulo x^n */ /* Get the inverse of h, and store the result in f. */ void polyinv(poly &h, int n, poly &f) { pretreat(h, n); static poly_t tmp; fill(f, f + n + n, 0), f[0] = qpow(h[0], MOD - 2); for(int t = 2; t <= n; t <<= 1) { copy(h, h + t, tmp), fill(tmp + t, tmp + t + t, 0); FNTT(f, t + t, 1), FNTT(tmp, t + t, 1); for(int i = 0; i != t + t; ++i) f[i] = qmul(f[i], qsub(2, qmul(f[i], tmp[i]))); FNTT(f, t + t, -1); fill(f + t, f + t + t, 0); } } /* Get the derivative of h, and store the result in f. */ void polyder(poly & h, int n, poly & f) { for(int i = 1; i != n; ++i) f[i - 1] = qmul(h[i], i); f[n - 1] = 0; } /* Get the integral of h, and store the result in f. */ void polyint(poly &h, int n, poly &f, int C = 0) { static int inv[MAXN]; if(inv[0] == 0) { inv[1] = 1; for(int i = 2; i < MAXN; ++i) inv[i] = qmul(inv[MOD % i], (MOD - MOD / i)); inv[0] = -1; } for(int i = n - 1; i != 0; --i) f[i] = qmul(h[i - 1], inv[i]); /* or inv[i] = qpow(i, MOD - 2) */ f[0] = C; /* constant C */ } /* Get the logarithm of h, and store the result in f. */ void polylog(poly &h, int n, poly & f) { pretreat(h, n); static poly_t tmp; polyder(h, n, tmp), polyinv(h, n, f); fill(tmp + n, tmp + n + n, 0); FNTT(tmp, n + n, 1), FNTT(f, n + n, 1); for(int i = 0; i != n + n; ++i) tmp[i] = qmul(tmp[i], f[i]); FNTT(tmp, n + n, -1); polyint(tmp, n, f); } /* Get the exponent of h, and store the result in f. */ void polyexp(poly & h, int n, poly & f) { pretreat(h, n); static poly_t tmp; fill(f, f + n + n, 0), f[0] = 1; for(int t = 2; t <= n; t <<= 1) { polylog(f, t, tmp); tmp[0] = qsub(qadd(h[0], 1), tmp[0]); for(int i = 1; i != t; ++i) tmp[i] = qsub(h[i], tmp[i]); fill(tmp + t, tmp + t + t, 0); FNTT(f, t + t, 1), FNTT(tmp, t + t, 1); for(int i = 0; i != t + t; ++i) f[i] = qmul(f[i], tmp[i]); FNTT(f, t + t, -1); fill(f + t, f + t + t, 0); } } /* 常数项为 1 的多项式 */ void polypow1(poly &h, int n, int k1, int k2, poly &f) { static poly_t tmp; polylog(h, n, tmp); for(int i = 0; i != n; ++i) tmp[i] = qmul(tmp[i], k1); polyexp(tmp, n, f); } /* 常数项非 {0, 1} 的多项式 */ void polypow2(poly &h, int n, int k1, int k2, poly &f) { for(int i = 0, invh = qpow(h[0], MOD - 2); i != n; ++i) f[i] = qmul(h[i], invh); polypow1(h, n, k1, k2, f); for(int i = 0, hk = qpow(h[0], k2); i != n; ++i) f[i] = qmul(f[i], hk); } /* Get the power k of h, and store the result in f. k1 = k mod MOD, k2 = k mod (MOD - 1), k = true k 常数项为0 */ void polypow(poly &h, int n, int k1, int k2, int k, poly &f) { int t = 0; for(; t < n && h[t] == 0; ++t); if(1LL * k * t >= n) fill(f, f + n, 0); else { static poly_t tmp; for(int i = 0; i + t != n; ++i) tmp[i] = h[i + t]; fill(tmp + n - t, tmp + n, 0); polypow2(tmp, n, k1, k2, f); for(int i = n - 1; i >= k * t; --i) f[i] = f[i - k * t]; fill(f, f + k * t, 0); } }
最后那个有递归关系