洛谷 P6667 [清华集训2016] 如何优雅地求和

洛谷传送门

点值不好搞。考虑把它搞成系数一类的东西。

由二项式反演，\(f(x) = \sum\limits_{i = 0}^x \binom{x}{i} b_i \Leftrightarrow b_i = \sum\limits_{j = 0}^i \binom{i}{j} (-1)^{i - j} f(j)\)。

然后我们要求：

\[\sum\limits_{k = 0}^n \sum\limits_{i = 0}^m s_i \binom{k}{i} \binom{n}{k} x^k (1 - x)^{n - k} \]

然后把 \(\binom{n}{k} \binom{k}{i}\) 拆成 \(\binom{n}{i} \binom{n - i}{k - i}\)，利用二项式定理合并即可。

\(b_i\) 可以一遍加法卷积求。

code

// Problem: P6667 [清华集训2016] 如何优雅地求和
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P6667
// Memory Limit: 128 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

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

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

const int maxn = 100100;
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, X, a[maxn], r[maxn], b[maxn], fac[maxn], ifac[maxn], inv[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;
}

void solve() {
	scanf("%lld%lld%lld", &n, &m, &X);
	fac[0] = 1;
	for (int i = 1; i <= m; ++i) {
		fac[i] = fac[i - 1] * i % mod;
	}
	ifac[m] = qpow(fac[m], mod - 2);
	for (int i = m - 1; ~i; --i) {
		ifac[i] = ifac[i + 1] * (i + 1) % mod;
	}
	inv[1] = 1;
	for (int i = 2; i <= m; ++i) {
		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	}
	for (int i = 0; i <= m; ++i) {
		scanf("%lld", &a[i]);
	}
	int k = 0;
	while ((1 << k) <= (m + 1) * 2) {
		++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 <= m; ++i) {
		A[i] = ifac[i] * a[i] % mod;
		B[i] = ifac[i] * ((i & 1) ? mod - 1 : 1) % mod;
	}
	poly C = A * B;
	for (int i = 0; i <= m; ++i) {
		b[i] = C[i] * fac[i] % mod;
	}
	ll c = 1, ans = 0, pw = 1;
	for (int i = 0; i <= m; ++i) {
		ans = (ans + b[i] * c % mod * pw % mod) % mod;
		c = c * (n - i) % mod * inv[i + 1] % mod;
		pw = pw * X % mod;
	}
	printf("%lld\n", ans);
}

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