AtCoder Beginner Contest 267 Ex Odd Sum

洛谷传送门

AtCoder 传送门

直接暴力跑背包的复杂度太高了，考虑优化。

发现值域很小，对值域从小到大跑背包。设 \(f_i\) 为用奇数个数凑出和为 \(i\) 的方案数，相对地 \(g_i\) 是用偶数个数。设当前枚举到的值为 \(x\)，数量为 \(c_x\)，那么我们要做的就是：

• \(f_i \times \binom{c_x}{d} \to f_{i + dx}, d \mid 2\)；
• \(f_i \times \binom{c_x}{d} \to g_{i + dx}, d \nmid 2\)；
• \(g_i \times \binom{c_x}{d} \to f_{i + dx}, d \nmid 2\)；
• \(g_i \times \binom{c_x}{d} \to g_{i + dx}, d \mid 2\)。

至此可以观察出卷积形式，NTT 优化。

每次枚举 \(x\)，实际上要做 \(4\) 次卷积，\(12\) 次 NTT。直接跑是 \(O(120 m \log m)\) 的，还是过不去。考虑做一些上界的优化，\(f_i, g_i\) 的下标上界设置为 \(\sum\limits_{j=1}^x j \times c_j\)，就可以通过了。

code

// Problem: Ex - Odd Sum
// Contest: AtCoder - NEC Programming Contest 2022 (AtCoder Beginner Contest 267)
// URL: https://atcoder.jp/contests/abc267/tasks/abc267_h
// Memory Limit: 1024 MB
// Time Limit: 4000 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 = 2200100;
const ll mod = 998244353;
const ll 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[15], fac[maxn], ifac[maxn], f[maxn], g[maxn], r[maxn];

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;
	}
}

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;
			}
		}
	}
	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);
	ll inv = qpow(n, mod - 2);
	for (int i = 0; i < n; ++i) {
		a[i] = a[i] * inv % mod;
	}
	return a;
}

void solve() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1, x; i <= n; ++i) {
		scanf("%d", &x);
		++a[x];
	}
	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;
	}
	f[0] = 1;
	ll s = 0;
	for (int x = 1; x <= 10; ++x) {
		int k = 0;
		while ((1 << k) <= min(s, m) + a[x] * x) {
			++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 <= min(s, m); ++i) {
			A[i] = f[i];
		}
		for (int i = 0; i <= a[x] * x; ++i) {
			B[i] = ((i % x || (i / x) % 2) ? 0 : C(a[x], i / x));
		}
		poly r1 = A * B;
		A = poly(1 << k);
		B = poly(1 << k);
		for (int i = 0; i <= min(s, m); ++i) {
			A[i] = g[i];
		}
		for (int i = 0; i <= a[x] * x; ++i) {
			B[i] = ((i % x || (i / x) % 2 == 0) ? 0 : C(a[x], i / x));
		}
		poly r2 = A * B;
		A = poly(1 << k);
		B = poly(1 << k);
		for (int i = 0; i <= min(s, m); ++i) {
			A[i] = f[i];
		}
		for (int i = 0; i <= a[x] * x; ++i) {
			B[i] = ((i % x || (i / x) % 2 == 0) ? 0 : C(a[x], i / x));
		}
		poly r3 = A * B;
		A = poly(1 << k);
		B = poly(1 << k);
		for (int i = 0; i <= min(s, m); ++i) {
			A[i] = g[i];
		}
		for (int i = 0; i <= a[x] * x; ++i) {
			B[i] = ((i % x || (i / x) % 2) ? 0 : C(a[x], i / x));
		}
		poly r4 = A * B;
		s += a[x] * x;
		for (int i = 0; i <= min(s, m); ++i) {
			f[i] = (r1[i] + r2[i]) % mod;
			g[i] = (r3[i] + r4[i]) % mod;
		}
	}
	printf("%lld\n", g[m]);
}

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