AtCoder Beginner Contest 241 Ex Card Deck Score

洛谷传送门

AtCoder 传送门

答案即为：

\[\sum\limits_c \prod\limits_{i = 1}^n [c_i \le b_i] a_i^{c_i} \]

考虑生成函数，设 \(F_i(x) = \sum\limits_{j = 0}^{b_i} (a_i x)^j\)。那么答案即为 \([x^m] \prod\limits_{i = 1}^n F_i(x)\)。

考虑 \(F_i(x) = \sum\limits_{j = 0}^{b_i} (a_i x)^j = \frac{1 - (a_i x)^{b_i + 1}}{1 - a_i x}\)。分子可以 \(O(2^n)\) 枚举乘了哪一项，这样可以知道分母需要贡献 \(x\) 的多少次方。我们重点关注分母。设现在要求分母的 \(x^q\) 次项。

有一个常见套路是设 \(\sum\limits_{i = 1}^n \frac{f_i}{1 - a_i x} = \frac{1}{\prod\limits_{i = 1}^n (1 - a_i x)}\)。因为 \(\sum\limits_{i = 1}^n \frac{f_i}{1 - a_i x} = \frac{\sum\limits_{i = 1}^n f_i \prod\limits_{j \ne i} (1 - a_j x)}{\prod\limits_{i = 1}^n (1 - a_i x)}\) 所以可以列出 \(n\) 个关于 \(f_i\) 的方程，可以高斯消元求解。

那么 \([x^q] \frac{1}{\prod\limits_{i = 1}^n (1 - a_i x)} = [x^q] \sum\limits_{i = 1}^n \frac{f_i}{1 - a_i x} = \sum\limits_{i = 1}^n f_i ([x^q] \sum\limits_{j \ge 0} (a_i x)^j) = \sum\limits_{i = 1}^n f_i a_i^q\)。

总时间复杂度 \(O(n (2^n \log m + n^2))\)。

code

// Problem: Ex - Card Deck Score
// Contest: AtCoder - AtCoder Beginner Contest 241（Sponsored by Panasonic）
// URL: https://atcoder.jp/contests/abc241/tasks/abc241_h
// Memory Limit: 1024 MB
// Time Limit: 3000 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 = 20;
const ll mod = 998244353;

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], b[maxn], c[maxn];

typedef vector<ll> poly;

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

ll f[maxn][maxn];
poly F[maxn];

void solve() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld%lld", &a[i], &b[i]);
		c[i] = qpow(a[i] % mod, b[i] + 1);
	}
	for (int i = 1; i <= n; ++i) {
		F[i] = poly(1, 1);
		for (int j = 1; j <= n; ++j) {
			if (i != j) {
				poly p;
				p.pb(1);
				p.pb(mod - a[j]);
				F[i] = F[i] * p;
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			f[i][j] = F[j][i - 1];
		}
		f[i][n + 1] = (i == 1);
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = i; j <= n; ++j) {
			if (f[j][i]) {
				swap(f[i], f[j]);
				break;
			}
		}
		ll t = qpow(f[i][i], mod - 2);
		for (int j = i; j <= n + 1; ++j) {
			f[i][j] = f[i][j] * t % mod;
		}
		for (int j = 1; j <= n; ++j) {
			if (i == j) {
				continue;
			}
			ll t = (mod - f[j][i]) % mod;
			for (int k = i + 1; k <= n + 1; ++k) {
				f[j][k] = (f[j][k] + t * f[i][k]) % mod;
			}
		}
	}
	ll ans = 0;
	for (int S = 0; S < (1 << n); ++S) {
		ll p = 1, q = 0;
		for (int i = 1; i <= n; ++i) {
			if (S & (1 << (i - 1))) {
				p = (mod - p * c[i] % mod) % mod;
				q += b[i] + 1;
			}
		}
		if (q > m) {
			continue;
		}
		for (int i = 1; i <= n; ++i) {
			ans = (ans + p * f[i][n + 1] % mod * qpow(a[i], m - q)) % mod;
		}
	}
	printf("%lld\n", ans);
}

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