【校内训练2019-04-03】星际穿越

最新推荐文章于 2022-04-28 20:42:35 发布

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最新推荐文章于 2022-04-28 20:42:35 发布

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分类专栏：【类型】做题记录【算法】生成函数【算法】动态规划【算法】容斥原理【算法】组合数学

本文链接：https://blog.youkuaiyun.com/qq_39972971/article/details/88992786

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本文介绍了一种使用多项式求逆和快速傅里叶变换优化的算法来解决极长上升序列计数的问题，该算法的时间复杂度为O(NLogN)。

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【思路要点】

考虑 $r = 1$ ，问题要求将排列分成若干段长度为 $k$ 的极长上升序列，这里假设 $N$ 是 $k$ 的倍数， $N$ 不为 $k$ 的倍数时只需要多一些细节处理。
若不考虑极长的限制，这里的排列数显然为 $N!k!Nk\frac{N!}{k!^{\frac{N}{k}}}$ ，那么，我们可以用容斥原理计算考虑极长限制的方案数，即记 $dp_i$ 表示第 $1∼i1\sim i$ 段的分割方式数除去 $N!$ 后的数值，则有
$dpi=−∑j=0i−1dpj((i−j)×k)!dp_{i}=-\sum_{j=0}^{i-1}\frac{dp_j}{((i-j)\times k)!}$
可以用多项式求逆优化上式，时间复杂度 $O (N L o g N)$ 。
由上面的算法不难得到 $r > 1$ 的转移
$dpi=−∑j=0i−1dpj((i−j)×k)!rdp_{i}=-\sum_{j=0}^{i-1}\frac{dp_j}{((i-j)\times k)!^r}$
同样可以用多项式求逆优化上式，时间复杂度 $O (N L o g N)$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2097152;
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("");
}
namespace Poly {
	const int MAXN = 2097152;
	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 invn[MAXN], tmpa[MAXN], tmpb[MAXN];
	int N, Log, home[MAXN]; bool initialized;
	int forward[MAXN], bckward[MAXN], inv[LOG];
	void init() {
		initialized = true;
		forward[0] = bckward[0] = inv[0] = invn[1] = 1;
		for (int len = 2, lg = 1; len <= MAXN; len <<= 1, lg++)
			inv[lg] = power(len, P - 2);
		for (int i = 2; i < MAXN; i++)
			invn[i] = (P - 1ll * (P / i) * invn[P % i] % P) % P;
		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 (int i = 0; i < a.size(); i++)
			tmpa[i] = a[i];
		for (int i = a.size(); i < N; i++)
			tmpa[i] = 0;
		for (int 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 (int i = 0; i < a.size(); i++)
			tmpa[i] = a[i];
		for (int i = a.size(); i < N; i++)
			tmpa[i] = 0;
		for (int 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 getder(vector <int> &a, vector <int> &b) {
		assert(a.size() >= 1);
		if (a.size() == 1) {
			b.clear();
			b.resize(1);
		} else {
			b.resize(a.size() - 1);
			for (unsigned i = 0; i < b.size(); i++)
				b[i] = (i + 1ll) * a[i + 1] % P;
		}
	}
	void getint(vector <int> &a, vector <int> &b) {
		b.resize(a.size() + 1), b[0] = 0;
		for (unsigned i = 0; i < a.size(); i++)
			b[i + 1] = 1ll * invn[i + 1] * a[i] % P;
	}
	void getlog(vector <int> &a, vector <int> &b) {
		assert(a.size() >= 1), assert(a[0] == 1);
		vector <int> da, inva, db;
		getder(a, da), getinv(a, inva);
		times(da, inva, db), getint(db, b);
		b.resize(a.size());
	}
	void getexp(vector <int> &a, vector <int> &b) {
		assert(a.size() >= 1), assert(a[0] == 0);
		b.clear(), b.push_back(1);
		while (b.size() < a.size()) {
			vector <int> lnb, res;
			b.resize(b.size() * 2), getlog(b, lnb);
			for (unsigned i = 0; i < lnb.size(); i++)
				if (i == 0) lnb[i] = (P + 1 + a[i] - lnb[i]) % P;
				else if (i < a.size()) lnb[i] = (P + a[i] - lnb[i]) % P;
				else lnb[i] = (P - lnb[i]) % P;
			times(lnb, b, res);
			res.resize(b.size());
			swap(res, b);
		}
		b.resize(a.size());
	}
}
int n, m, r, k, s[MAXN], dp[MAXN];
int fac[MAXN], inv[MAXN], coef[MAXN];
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 getc(int x, int y) {
	if (y > x) return 0;
	else return 1ll * fac[x] * inv[y] % P * inv[x - y] % P;
}
void init(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
		fac[i] = 1ll * fac[i - 1] * i % P;
	inv[n] = power(fac[n], P - 2);
	for (int i = n - 1; i >= 0; i--)
		inv[i] = inv[i + 1] * (i + 1ll) % P;
}
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int main() {
	freopen("interstellar.in", "r", stdin);
	freopen("interstellar.out", "w", stdout);
	read(m), read(r), read(k), init(m);
	n = (m - 1) / k + 1;
	vector <int> invr, res;
	for (int i = 0; i <= n - 1; i++)
		invr.push_back(power(inv[i * k], r));
	Poly :: init(), Poly :: getinv(invr, res);
	int ans = 0;
	for (int i = 0; i <= n - 1; i++)
		update(ans, P - 1ll * res[i] * power(inv[m - i * k], r) % P);
	ans = 1ll * ans * power(fac[m], r) % P;
	if (n & 1) writeln((P - ans) % P);
	else writeln(ans);
	return 0;
}