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【题目链接】
【思路要点】
- 记 表示从 号点走到 号点的方案数,有
- 考虑容斥原理,我们强制一部分格子没有被踩过,计算在剩余格子中行走的方案数,乘上容斥系数计入答案。
- 记 表示从 走到 ,强制 未被踩过,带有容斥系数的方案数, 表示 时的答案,则有
- 记 ,则有 ,即
- 设 ,解得
- 因此 。
- 令 ,则有
- 进而
- 因此,记 , 均为次数 的多项式,且 ,其常数项为 。进而 为线性递推数列,其递推式即为 。
- 记 ,由于 均为递推式与 相同的线性递推数列,它们同样可以写成 的形式,其中 为次数 的多项式。
- 令 ,那么
- 可以发现, 也为线性递推数列,采用多项式乘法、带余除法优化线性递推即可。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e5 + 5; const int P = 998244353; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int n, m, p, q, t; inline int pls(const int &a, const int &b) {return (a + b >= P) ? (a + b - P) : (a + b); } inline int mns(const int &a, const int &b) {return (a - b >= 0) ? (a - b) : (a - b + P); } inline int mul(const int &a, const int &b) {return 1ll * a * b % P; } int power(int x, int y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return mul(tmp, tmp); else return mul(mul(tmp, tmp), x); } struct INT {int r, i; }; INT operator * (const INT &a, const int &b) {return (INT) {mul(a.r, b), mul(a.i, b)}; } INT operator + (const INT &a, const INT &b) {return (INT) {pls(a.r, b.r), pls(a.i, b.i)}; } INT operator - (const INT &a, const INT &b) {return (INT) {mns(a.r, b.r), mns(a.i, b.i)}; } INT operator * (const INT &a, const INT &b) {return (INT) {pls(mul(a.r, b.r), mul(mul(a.i, b.i), t)), pls(mul(a.r, b.i), mul(b.r, a.i))}; } namespace Poly { const int MAXN = 262144; const int P = 998244353; const int LOG = 25; const int G = 3; int power(int x, int y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return 1ll * tmp * tmp % P; else return 1ll * tmp * tmp % P * x % P; } int N, Log, tmpa[MAXN], tmpb[MAXN], home[MAXN]; bool initialized; int forward[MAXN], bckward[MAXN], inv[LOG]; void init() { initialized = true; forward[0] = bckward[0] = inv[0] = 1; for (int len = 2, lg = 1; len <= MAXN; len <<= 1, lg++) inv[lg] = power(len, P - 2); int delta = power(G, (P - 1) / MAXN); for (int i = 1; i < MAXN; i++) forward[i] = bckward[MAXN - i] = 1ll * forward[i - 1] * delta % P; } void NTTinit() { for (int i = 0; i < N; i++) { int ans = 0, tmp = i; for (int j = 1; j <= Log; j++) { ans <<= 1; ans += tmp & 1; tmp >>= 1; } home[i] = ans; } } void NTT(int *a, int mode) { assert(initialized); for (int i = 0; i < N; i++) if (home[i] < i) swap(a[i], a[home[i]]); int *g; if (mode == 1) g = forward; else g = bckward; for (int len = 2, lg = 1; len <= N; len <<= 1, lg++) { for (int i = 0; i < N; i += len) { for (int j = i, k = i + len / 2; k < i + len; j++, k++) { int tmp = a[j]; int tnp = 1ll * a[k] * g[MAXN / len * (j - i)] % P; a[j] = (tmp + tnp > P) ? (tmp + tnp - P) : (tmp + tnp); a[k] = (tmp - tnp < 0) ? (tmp - tnp + P) : (tmp - tnp); } } } if (mode == -1) { for (int i = 0; i < N; i++) a[i] = 1ll * a[i] * inv[Log] % P; } } void times(vector <int> &a, vector <int> &b, vector <int> &c) { assert(a.size() >= 1), assert(b.size() >= 1); int goal = a.size() + b.size() - 1; N = 1, Log = 0; while (N < goal) { N <<= 1; Log++; } for (unsigned i = 0; i < a.size(); i++) tmpa[i] = a[i]; for (int i = a.size(); i < N; i++) tmpa[i] = 0; for (unsigned i = 0; i < b.size(); i++) tmpb[i] = b[i]; for (int i = b.size(); i < N; i++) tmpb[i] = 0; NTTinit(); NTT(tmpa, 1); NTT(tmpb, 1); for (int i = 0; i < N; i++) tmpa[i] = 1ll * tmpa[i] * tmpb[i] % P; NTT(tmpa, -1); c.resize(goal); for (int i = 0; i < goal; i++) c[i] = tmpa[i]; } void timesabb(vector <int> &a, vector <int> &b, vector <int> &c) { assert(a.size() >= 1), assert(b.size() >= 1); int goal = a.size() + b.size() * 2 - 2; N = 1, Log = 0; while (N < goal) { N <<= 1; Log++; } for (unsigned i = 0; i < a.size(); i++) tmpa[i] = a[i]; for (int i = a.size(); i < N; i++) tmpa[i] = 0; for (unsigned i = 0; i < b.size(); i++) tmpb[i] = b[i]; for (int i = b.size(); i < N; i++) tmpb[i] = 0; NTTinit(); NTT(tmpa, 1); NTT(tmpb, 1); for (int i = 0; i < N; i++) tmpa[i] = 1ll * tmpa[i] * tmpb[i] % P * tmpb[i] % P; NTT(tmpa, -1); c.resize(goal); for (int i = 0; i < goal; i++) c[i] = tmpa[i]; } void getinv(vector <int> &a, vector <int> &b) { assert(a.size() >= 1), assert(a[0] != 0); b.clear(), b.push_back(power(a[0], P - 2)); while (b.size() < a.size()) { vector <int> c, ta = a; ta.resize(b.size() * 2); timesabb(ta, b, c); b.resize(b.size() * 2); for (unsigned i = 0; i < b.size(); i++) b[i] = (2ll * b[i] - c[i] + P) % P; } b.resize(a.size()); } void getdiv(vector <int> &a, vector <int> &b, vector <int> &q) { a.resize(max(a.size(), b.size())); reverse(a.begin(), a.end()); reverse(b.begin(), b.end()); vector <int> invb; getinv(b, invb); times(a, invb, q), q.resize(a.size() - b.size() + 1); reverse(a.begin(), a.end()); reverse(b.begin(), b.end()); reverse(q.begin(), q.end()); } void getmod(vector <int> &a, vector <int> &b, vector <int> &r) { vector <int> q, p; getdiv(a, b, q); times(b, q, p), r.clear(); for (unsigned i = 0; i < b.size() - 1; i++) r.push_back((a[i] - p[i] >= 0) ? (a[i] - p[i]) : (a[i] - p[i] + P)); } void NTT(INT *a, int mode) { assert(initialized); for (int i = 0; i < N; i++) if (home[i] < i) swap(a[i], a[home[i]]); int *g; if (mode == 1) g = forward; else g = bckward; for (int len = 2, lg = 1; len <= N; len <<= 1, lg++) { for (int i = 0; i < N; i += len) { for (int j = i, k = i + len / 2; k < i + len; j++, k++) { INT tmp = a[j]; INT tnp = a[k] * g[MAXN / len * (j - i)]; a[j] = tmp + tnp; a[k] = tmp - tnp; } } } if (mode == -1) { for (int i = 0; i < N; i++) a[i] = a[i] * inv[Log]; } } void times(vector <INT> &a, vector <INT> &b, vector <INT> &c) { assert(a.size() >= 1), assert(b.size() >= 1); int goal = a.size() + b.size() - 1; N = 1, Log = 0; while (N < goal) { N <<= 1; Log++; } static INT tmpa[MAXN], tmpb[MAXN]; for (unsigned i = 0; i < a.size(); i++) tmpa[i] = a[i]; for (int i = a.size(); i < N; i++) tmpa[i] = (INT) {0, 0}; for (unsigned i = 0; i < b.size(); i++) tmpb[i] = b[i]; for (int i = b.size(); i < N; i++) tmpb[i] = (INT) {0, 0}; NTTinit(); NTT(tmpa, 1); NTT(tmpb, 1); for (int i = 0; i < N; i++) tmpa[i] = tmpa[i] * tmpb[i]; NTT(tmpa, -1); c.resize(goal); for (int i = 0; i < goal; i++) c[i] = tmpa[i]; } } namespace LinearSequence { const int MAXN = 262144; const int P = 998244353; vector <int> mod, inv; int power(int x, long long y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return 1ll * tmp * tmp % P; else return 1ll * tmp * tmp % P * x % P; } void times(vector <int> &a, vector <int> &b, vector <int> &res) { vector <int> tmp, q, p; Poly :: times(a, b, tmp); if (tmp.size() < mod.size()) { res = tmp; return; } reverse(tmp.begin(), tmp.end()); Poly :: times(tmp, inv, q); q.resize(tmp.size() - inv.size() + 1); reverse(tmp.begin(), tmp.end()); reverse(q.begin(), q.end()); Poly :: times(mod, q, p), res.clear(); for (unsigned i = 0; i < mod.size() - 1; i++) res.push_back((tmp[i] - p[i] >= 0) ? (tmp[i] - p[i]) : (tmp[i] - p[i] + P)); } vector <int> work(ll n) { if (n == 0) { vector <int> ans; ans.push_back(1); return ans; } vector <int> tmp = work(n / 2), tnp = tmp, ans; if (n & 1) tmp.insert(tmp.begin(), 0); times(tmp, tnp, ans); return ans; } int query(long long n, vector <int> a, vector <int> h) { assert(h.size() >= a.size()); if (a.size() == 1) return 1ll * power(a[0], n) * h[0] % P; mod = a, reverse(mod.begin(), mod.end()); for (unsigned i = 0; i < mod.size(); i++) mod[i] = (mod[i] == 0) ? 0 : (P - mod[i]); mod.push_back(1); reverse(mod.begin(), mod.end()); Poly :: getinv(mod, inv); reverse(mod.begin(), mod.end()); vector <int> res = work(n); int ans = 0; for (unsigned i = 0; i < res.size(); i++) ans = (ans + 1ll * res[i] * h[i]) % P; return ans; } } INT alpha, beta; INT power(INT x, int y) { if (y == 0) return (INT) {1, 0}; INT tmp = power(x, y / 2); if (y % 2 == 0) return tmp * tmp; else return tmp * tmp * x; } vector <INT> work(int l, int r) { if (l == r) { vector <INT> ans; ans.push_back((INT) {1, 0}); ans.push_back((INT) {0, 0} - power(alpha, l) * power(beta, m - l)); return ans; } int mid = (l + r) / 2; vector <INT> a = work(l, mid), b = work(mid + 1, r), ans; Poly :: times(a, b, ans); return ans; } int ways[MAXN]; int main() { read(n), read(m), read(p), read(q); t = pls(mul(p, p), mul(4, q)); alpha = (INT) {p, 1} * ((P + 1) / 2); beta = (INT) {p, -1} * ((P + 1) / 2); Poly :: init(); vector <INT> tmp = work(0, m); vector <int> Q; for (int i = 0; i <= m + 1; i++) { Q.push_back(tmp[i].r); assert(tmp[i].i == 0); } ways[1] = 1; for (int i = 2; i <= m + 2; i++) ways[i] = pls(mul(ways[i - 1], p), mul(ways[i - 2], q)); vector <int> f, g; for (int i = 0; i <= m + 1; i++) { f.push_back(power(mul(q, ways[i + 1]), m)); g.push_back(power(pls(mul(q, ways[i]), ways[i + 1]), m)); } vector <int> F, G, invF, Ans; Poly :: times(Q, f, F); Poly :: times(Q, g, G); for (int i = m + 1; i >= 0; i--) F[i] = pls(Q[i], (i >= 2) ? F[i - 2] : 0); Poly :: getinv(F, invF); Poly :: times(G, invF, Ans); Ans.resize(m + 2); for (int i = 0; i <= m; i++) F[i] = mns(0, F[i + 1]); F.resize(m + 1); writeln(LinearSequence :: query(n, F, Ans)); return 0; }