AtCoder Beginner Contest 212 H Nim Counting

洛谷传送门

AtCoder 传送门

牛逼题……

考虑如果定义 \(x^a \times x^b = x^{a \oplus b}\)，设 \(f(x) = \sum\limits_{i=1}^k x^{a_i}\)，那么题目就是求，\(\forall w > 0, \sum\limits_{i=1}^n (f(x))^i\)，指数为 \(w\) 的系数之和。

考虑 FWT，把每一项变成 \(\sum\limits_{i=1}^n x_i\) 再逆回去。

code

// Problem: H - Nim Counting
// Contest: AtCoder - AtCoder Beginner Contest 212
// URL: https://atcoder.jp/contests/abc212/tasks/abc212_h
// Memory Limit: 1024 MB
// Time Limit: 2000 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 = 200100;
const int U = (1 << 16);
const ll mod = 998244353;
const ll inv2 = (mod + 1) / 2;

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];

void FWT(ll *a, ll op) {
	for (int k = 1; k < U; k <<= 1) {
		for (int i = 0; i < U; i += (k << 1)) {
			for (int j = 0; j < k; ++j) {
				ll x = a[i + j], y = a[i + j + k];
				a[i + j] = (x + y) * op % mod;
				a[i + j + k] = (x - y + mod) % mod * op % mod;
			}
		}
	}
}

void solve() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= m; ++i) {
		scanf("%lld", &a[i]);
		++b[a[i]];
	}
	FWT(b, 1);
	for (int i = 0; i < U; ++i) {
		if (b[i] == 1) {
			b[i] = n;
			continue;
		}
		b[i] = ((qpow(b[i], n + 1) + mod - 1) % mod * qpow((b[i] + mod - 1) % mod, mod - 2) % mod + mod - 1) % mod;
	}
	FWT(b, inv2);
	ll ans = 0;
	for (int i = 1; i < U; ++i) {
		ans = (ans + b[i]) % mod;
	}
	printf("%lld\n", ans);
}

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