1 条题解
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0

#include <bits/stdc++.h> #define lg2(x) (31 - __builtin_clz(x)) typedef long long ll; const int N = 530000, mod = 998244353, half_mod = (mod + 1) / 2, root = 31, iv3 = (mod + 1) / 3; typedef int vec[N], *pvec; inline int & reduce(int &x) {return x += x >> 31 & mod;} ll PowerMod(ll a, int n, ll c = 1) {for (; n; n >>= 1, a = a * a % mod) if (n & 1) c = c * a % mod; return c;} namespace Poly { int l, n; vec rev, x, y; void NTT_init(int len) { if (l == len) return; n = 1 << (l = len); ll g = PowerMod(root, 1 << (23 - l)); *x = 1, *rev = 0; for (int i = 1; i < n; ++i) x[i] = x[i - 1] * g % mod, rev[i] = rev[i >> 1] >> 1 | (i & 1) << (l - 1); } void DNTT(int *d, int *t) { int i, *j, *k, len = 1, delta = n, R; for (i = 0; i < n; ++i) t[rev[i]] = d[i]; for (i = 0; i < l; ++i) { delta >>= 1; for (k = x, j = y; j < y + len; k += delta, ++j) *j = *k; for (j = t; j < t + n; j += len << 1) for (k = j; k < j + len; ++k) R = (ll)y[k - j] * k[len] % mod, k[len] = (*k - R < 0 ? *k - R + mod : *k - R), *k = (*k + R >= mod ? *k + R - mod : *k + R); len <<= 1; } } inline void IDNTT(int *d, int *t) { ll iv = mod - (mod - 1) / n; DNTT(d, t), std::reverse(t + 1, t + n); for (int i = 0; i < n; ++i) t[i] = t[i] * iv % mod; } vec B1, B2, B3; void Mul(int deg, pvec a, pvec b, pvec c) { if (!deg) {*c = (ll)*a * *b % mod; return;} NTT_init(lg2(deg) + 1); DNTT(a, c), DNTT(b, B1); for (int i = 0; i < n; ++i) B1[i] = (ll)B1[i] * c[i] % mod; IDNTT(B1, c); } void Inv(int deg, pvec a, pvec b) { int len, i; ll iv = half_mod; *b = PowerMod(*a, mod - 2), b[1] = 0, *B1 = *a, B1[1] = a[1]; for (len = 0; 1 << len < deg; ++len) { NTT_init(len + 2); memset(b + (n >> 1), 0, n << 1), DNTT(b, B2); memset(B1 + (n >> 1), 0, n << 1), DNTT(B1, B3); for (i = 0; i < n; ++i) reduce(B2[i] = B2[i] * (2ll - (ll)B2[i] * B3[i] % mod) % mod); DNTT(B2, B3), std::reverse(B3 + 1, B3 + n), iv = (iv >> 1) + half_mod; for (i = 0; i < n >> 1; ++i) b[i] = B3[i] * iv % mod; memcpy(B1 + i, a + i, n << 1); } } } int D; vec f, f2, f3; vec f_ntt, fs_ntt, fc_ntt; vec C0, C1, C2, C3; int main() { int i, i2 = 0, i3 = 0, len, n = 8; scanf("%d", &D); *f = f[1] = f[2] = 1, f[3] = 2; for (len = 2; 1 << len <= D; ++len, n <<= 1) { for (; i2 * 2 < n; ++i2) f2[i2 * 2] = f[i2]; for (; i3 * 3 < n; ++i3) f3[i3 * 3] = f[i3]; Poly::NTT_init(len + 2); Poly::DNTT(f, f_ntt); for (i = 0; i < Poly::n; ++i) fs_ntt[i] = (ll)f_ntt[i] * f_ntt[i] % mod, fc_ntt[i] = (ll)fs_ntt[i] * f_ntt[i] % mod; Poly::IDNTT(fs_ntt, C0); Poly::IDNTT(fc_ntt, C1); *C2 = *C3 = 1; for (i = 0; i < n - 1; ++i) reduce(C2[i + 1] = (ll)(f3[i] - C1[i]) * iv3 % mod), C3[i + 1] = (f2[i] + C0[i]) * (half_mod - 1ll) % mod; Poly::Inv(n, C3, C0); Poly::Mul(n * 2 - 1, C2, C0, f); memset(f + n, 0, n << 2); } printf("%d\n", f[D]); return 0; }
- 1
信息
- ID
- 4682
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 10
- 标签
- 递交数
- 1
- 已通过
- 1
- 上传者