AtCoder Beginner Contest 235 G Gardens

洛谷传送门

考虑不要求每个盒子至少放一个球，答案即为 \(\sum\limits_{i=0}^A \binom{n}{i} \times \sum\limits_{i=0}^B \binom{n}{i} \times \sum\limits_{i=0}^C \binom{n}{i}\)。

加上这个条件，考虑容斥，钦定 \(i\) 个盒子不能放球，其他随意，则答案为 \(\sum\limits_{i=0}^n (-1)^i \binom{n}{i} \sum\limits_{j=0}^A \binom{n-i}{j} \times \sum\limits_{j=0}^B \binom{n-i}{j} \times \sum\limits_{j=0}^C \binom{n-i}{j}\)。

似乎没有快速算 \(\sum\limits_{i=0}^m \binom{n}{i}\) 的方法，但是观察发现式子中 \(m\) 不变，\(n\) 每次减一，考虑递推。设 \(f_i = \sum\limits_{j=0}^A \binom{i}{j}\)，则根据 \(\binom{n}{m} = \binom{n-1}{m} + \binom{n-1}{m-1}\)，可得 \(f_i = \frac{f_{i+1} + \binom{i}{A}}{2}\)。按照相同的方法递推出 \(g_i = \sum\limits_{j=0}^B \binom{i}{j}\) 和 \(h_i = \sum\limits_{j=0}^C \binom{i}{j}\)，答案即为 \(\sum\limits_{i=0}^n (-1)^i \binom{n}{i} f_{n-i} g_{n-i} h_{n-i}\)。

code

/*

p_b_p_b txdy
AThousandSuns txdy
Wu_Ren txdy
Appleblue17 txdy

*/

#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 long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 10000100;
const int N = 10000000;
const ll mod = 998244353;
const ll inv2 = (mod + 1) / 2;

ll n, X, Y, Z, fac[maxn], ifac[maxn], f[maxn], g[maxn], h[maxn];

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;
}

inline ll C(ll n, ll m) {
	if (n < m || n < 0 || m < 0) {
		return 0;
	} else {
		return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
	}
}

void solve() {
	scanf("%lld%lld%lld%lld", &n, &X, &Y, &Z);
	fac[0] = 1;
	for (int i = 1; i <= N; ++i) {
		fac[i] = fac[i - 1] * i % mod;
	}
	ifac[N] = qpow(fac[N], mod - 2);
	for (int i = N - 1; ~i; --i) {
		ifac[i] = ifac[i + 1] * (i + 1) % mod;
	}
	for (int i = 0; i <= X; ++i) {
		f[n] = (f[n] + C(n, i)) % mod;
	}
	for (int i = n - 1; ~i; --i) {
		f[i] = (f[i + 1] + C(i, X)) % mod * inv2 % mod;
	}
	for (int i = 0; i <= Y; ++i) {
		g[n] = (g[n] + C(n, i)) % mod;
	}
	for (int i = n - 1; ~i; --i) {
		g[i] = (g[i + 1] + C(i, Y)) % mod * inv2 % mod;
	}
	for (int i = 0; i <= Z; ++i) {
		h[n] = (h[n] + C(n, i)) % mod;
	}
	for (int i = n - 1; ~i; --i) {
		h[i] = (h[i + 1] + C(i, Z)) % mod * inv2 % mod;
	}
	ll ans = 0;
	for (int i = 0; i <= n; ++i) {
		ans = (ans + ((i & 1) ? mod - 1 : 1) * C(n, i) % mod * f[n - i] % mod * g[n - i] % mod * h[n - i] % mod) % mod;
	}
	printf("%lld\n", ans);
}

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

[AtCoder 传送门](https://atcoder.jp/contests/abc235/tasks/abc235_g "AtCoder 传送门")