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AtCoder Beginner Contest 331 G Collect Them All

发布时间 2023-12-04 17:18:45作者: zltzlt

洛谷传送门

AtCoder 传送门

设数字 \(i\) 第一次拿到的时间为 \(t_i\)，所求即为 \(E(\max\limits_{i = 1}^m t_i)\)。

施 min-max 容斥，有：

\[\begin{aligned}E(\max\limits_{i = 1}^m t_i) & = \sum\limits_{T \subseteq [1, m] \land T \ne \varnothing} (-1)^{|T| - 1} E(\min\limits_{j \in T} t_j) \\ & = \sum\limits_{T \subseteq [1, m] \land T \ne \varnothing} (-1)^{|T| - 1} \frac{n}{\sum\limits_{j \in T} a_j} \end{aligned} \]

设 \(\sum\limits_{j \in T} a_j = k\) 的集合 \(T\) 的容斥系数 \((-1)^{|T| - 1}\) 之和为 \(f_k\)，那么：

\[ans = \sum\limits_{i = 1}^n \frac{n}{i} \times s_i \]

套路地，考虑生成函数 \(F(x) = -\prod\limits_{i = 1}^n (-x^{a_i} + 1)\)，那么 \(f_i = [x^i] F(x)\)。而 \(F(x)\) 可以分治 NTT 求出。具体就是递归求 \(\prod\limits_{i = l}^r (-x^{a_i} + 1)\)，然后得到 \([l, mid]\) 和 \([mid + 1, r]\) 的答案再合并，总时间复杂度就是 \(O(n \log^2 n)\)。

好像有先 exp 再求逆的 \(O(n \log n)\) 做法，但是我既不会 exp 也不会求逆，开摆。

code 
// Problem: G - Collect Them All
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2023（AtCoder Beginner Contest 331）
// URL: https://atcoder.jp/contests/abc331/tasks/abc331_g
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1000100;
const ll mod = 998244353, G = 3;

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

ll n, m, a[maxn], f[maxn], inv[maxn], r[maxn];

typedef vector<ll> poly;

inline poly NTT(poly a, int op) {
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		if (i < r[i]) {
			swap(a[i], a[r[i]]);
		}
	}
	for (int k = 1; k < n; k <<= 1) {
		ll wn = qpow(op == 1 ? G : qpow(G, mod - 2), (mod - 1) / (k << 1));
		for (int i = 0; i < n; i += (k << 1)) {
			ll w = 1;
			for (int j = 0; j < k; ++j, w = w * wn % mod) {
				ll x = a[i + j], y = w * a[i + j + k] % mod;
				a[i + j] = (x + y) % mod;
				a[i + j + k] = (x - y + mod) % mod;
			}
		}
	}
	if (op == -1) {
		ll inv = qpow(n, mod - 2);
		for (int i = 0; i < n; ++i) {
			a[i] = a[i] * inv % mod;
		}
	}
	return a;
}

inline poly operator * (poly a, poly b) {
	a = NTT(a, 1);
	b = NTT(b, 1);
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		a[i] = a[i] * b[i] % mod;
	}
	a = NTT(a, -1);
	return a;
}

inline poly mul(poly a, poly b) {
	int n = (int)a.size() - 1, m = (int)b.size() - 1, k = 0;
	while ((1 << k) <= n + m + 1) {
		++k;
	}
	for (int i = 1; i < (1 << k); ++i) {
		r[i] = (r[i >> 1] >> 1) | ((i & 1) << (k - 1));
	}
	poly A(1 << k), B(1 << k);
	for (int i = 0; i <= n; ++i) {
		A[i] = a[i];
	}
	for (int i = 0; i <= m; ++i) {
		B[i] = b[i];
	}
	poly res = A * B;
	res.resize(n + m + 1);
	return res;
}

inline poly dfs(int l, int r) {
	if (l == r) {
		poly res(a[l] + 1);
		res[0] = 1;
		res[a[l]] = mod - 1;
		return res;
	}
	int mid = (l + r) >> 1;
	return mul(dfs(l, mid), dfs(mid + 1, r));
}

void solve() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= m; ++i) {
		scanf("%lld", &a[i]);
	}
	inv[1] = 1;
	for (int i = 2; i <= n; ++i) {
		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	}
	poly res = dfs(1, m);
	ll ans = 0;
	for (int i = 1; i <= n; ++i) {
		ans = (ans + n * inv[i] % mod * res[i]) % mod;
	}
	ans = (mod - ans) % mod;
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}