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AtCoder Beginner Contest 309 Ex Simple Path Counting Problem

发布时间 2023-07-13 08:05:58作者: zltzlt

洛谷传送门

AtCoder 传送门

挺妙的题。

考虑 dp，\(f_{i, j}\) 表示走到 \((i, j)\) 方案数，则 \(f_{i, j} = [j > 1] f_{i - 1, j - 1} + f_{i - 1, j} + [j < m] f_{i - 1, j + 1}\)。

发现这个东西很像什么 \(x^{-1} + x^0 + x^1\) 的 \(n\) 次方，但是我们还要规定 \(f_{i, 0} = f_{i, m + 1} = 0\)，有点烦。

考虑把 dp 值沿着 \(f_{i, m + 1}\) 对称过去，并且 \(f_{i, j} + f_{i, 2m + 2 - j} = 0\)。这样的转移保证 \(f_{i, 0} = f_{i, m - 1} = 0\)。

于是我们做一个多项式 \(n\) 次幂即可。

code 
// Problem: Ex - Simple Path Counting Problem
// Contest: AtCoder - Denso Create Programming Contest 2023 (AtCoder Beginner Contest 309)
// URL: https://atcoder.jp/contests/abc309/tasks/abc309_h
// Memory Limit: 1024 MB
// Time Limit: 10000 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 = 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, B, a[maxn], b[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 qpow(poly t, ll p) {
	int n = (int)t.size();
	poly res(n);
	for (int i = 1; i <= A; ++i) {
		res[a[i]] = (res[a[i]] + 1) % mod;
		res[m * 2 + 2 - a[i]] = (res[m * 2 + 2 - a[i]] + mod - 1) % mod;
	}
	while (p) {
		if (p & 1) {
			res = res * t;
			for (int i = m * 2 + 2; i < n; ++i) {
				res[i % (m * 2 + 2)] = (res[i % (m * 2 + 2)] + res[i]) % mod;
				res[i] = 0;
			}
		}
		t = t * t;
		for (int i = m * 2 + 2; i < n; ++i) {
			t[i % (m * 2 + 2)] = (t[i % (m * 2 + 2)] + t[i]) % mod;
			t[i] = 0;
		}
		p >>= 1;
	}
	return res;
}

void solve() {
	scanf("%lld%lld%lld%lld", &n, &m, &A, &B);
	for (int i = 1; i <= A; ++i) {
		scanf("%lld", &a[i]);
	}
	for (int i = 1; i <= B; ++i) {
		scanf("%lld", &b[i]);
	}
	int k = 0;
	while ((1 << k) <= 4 * m + 4) {
		++k;
	}
	for (int i = 1; i < (1 << k); ++i) {
		r[i] = (r[i >> 1] >> 1) | ((i & 1) << (k - 1));
	}
	poly t(1 << k);
	t[m * 2 + 1] = t[0] = t[1] = 1;
	poly res = qpow(t, n - 1);
	ll ans = 0;
	for (int i = 1; i <= B; ++i) {
		ans = (ans + res[b[i]]) % mod;
	}
	printf("%lld\n", ans);
}

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