結果

問題 No.1239 Multiplication -2
ユーザー chocopuuchocopuu
提出日時 2020-09-25 23:44:09
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 4,218 bytes
コンパイル時間 2,236 ms
コンパイル使用メモリ 208,340 KB
実行使用メモリ 28,288 KB
最終ジャッジ日時 2024-06-28 08:23:23
合計ジャッジ時間 7,051 ms
ジャッジサーバーID
(参考情報)
judge3 / judge1
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テストケース

テストケース表示
入力 結果 実行時間
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testcase_00 WA -
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ソースコード

diff #

#include "bits/stdc++.h"
using namespace std;
//#include "atcoder/all"
//using namespace atcoder;
#define int long long
#define REP(i, n) for (int i = 0; i < (int)n; ++i)
#define RREP(i, n) for (int i = (int)n - 1; i >= 0; --i)
#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define RFOR(i, s, n) for (int i = (int)n - 1; i >= s; --i)
#define ALL(a) a.begin(), a.end()
#define IN(a, x, b) (a <= x && x < b)
template<class T>istream&operator >>(istream&is,vector<T>&vec){for(T&x:vec)is>>x;return is;}
template<class T>inline void out(T t){cout << t << "\n";}
template<class T,class... Ts>inline void out(T t,Ts... ts){cout << t << " ";out(ts...);}
template<class T>inline bool CHMIN(T&a,T b){if(a > b){a = b;return true;}return false;}
template<class T>inline bool CHMAX(T&a,T b){if(a < b){a = b;return true;}return false;}
constexpr int INF = 1e18;

template<int MOD> struct Fp {
	long long val;
	constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
		if (val < 0) val += MOD;
	}
	constexpr int getmod() { return MOD; }
	constexpr Fp operator - () const noexcept {
		return val ? MOD - val : 0;
	}
	constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
	constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
	constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
	constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
	constexpr Fp& operator += (const Fp& r) noexcept {
		val += r.val;
		if (val >= MOD) val -= MOD;
		return *this;
	}
	constexpr Fp& operator -= (const Fp& r) noexcept {
		val -= r.val;
		if (val < 0) val += MOD;
		return *this;
	}
	constexpr Fp& operator *= (const Fp& r) noexcept {
		val = val * r.val % MOD;
		return *this;
	}
	constexpr Fp& operator /= (const Fp& r) noexcept {
		long long a = r.val, b = MOD, u = 1, v = 0;
		while (b) {
			long long t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		val = val * u % MOD;
		if (val < 0) val += MOD;
		return *this;
	}
	constexpr bool operator == (const Fp& r) const noexcept {
		return this->val == r.val;
	}
	constexpr bool operator != (const Fp& r) const noexcept {
		return this->val != r.val;
	}
	friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
		return os << x.val;
	}
	friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
		if (n == 0) return 1;
		auto t = modpow(a, n / 2);
		t = t * t;
		if (n & 1) t = t * a;
		return t;
	}
};
 

template<class T> struct BiCoef {
	vector<T> fact_, inv_, finv_;
	constexpr BiCoef() {}
	constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
		init(n);
	}
	constexpr void init(int n) noexcept {
		fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
		int MOD = fact_[0].getmod();
		for(int i = 2; i < n; i++){
			fact_[i] = fact_[i-1] * i;
			inv_[i] = -inv_[MOD%i] * (MOD/i);
			finv_[i] = finv_[i-1] * inv_[i];
		}
	}
	constexpr T com(int n, int k) const noexcept {
		if (n < k || n < 0 || k < 0) return 0;
		return fact_[n] * finv_[k] * finv_[n-k];
	}
	constexpr T P(int n, int k) const noexcept {
		if (n < k || n < 0 || k < 0) return 0;
		return fact_[n] * finv_[n-k];
	}
	constexpr T fact(int n) const noexcept {
		if (n < 0) return 0;
		return fact_[n];
	}
	constexpr T inv(int n) const noexcept {
		if (n < 0) return 0;
		return inv_[n];
	}
	constexpr T finv(int n) const noexcept {
		if (n < 0) return 0;
		return finv_[n];
	}
};
 
// const int MOD = 1000000007;
const int MOD = 998244353;
using mint = Fp<MOD>;
BiCoef<mint> bc;
// bc.init(500050);

mint dp[200020][5][3];

signed main(){
	int N;
	cin >> N;
	vector<int>a(N);
	cin >> a;
	REP(i, N) {
		REP(j, 5) {
			int mul = j - 2;
			if(!IN(-2, mul * a[i], 2 + 1)) continue;
			dp[i + 1][a[i] * mul + 2][0] += dp[i][j][0];
			if(a[i] * mul == -2) {
				dp[i + 1][0][2] += dp[i][j][0] * modpow(mint(2), max(0ll ,N - i - 2));
			}
		}
		dp[i + 1][a[i] + 2][0] += modpow(mint(2), max(0ll, i - 1));
		if(a[i] + 2 == 0) dp[i + 1][0][2] += modpow(mint(2), max(0ll, i - 1)) * modpow(mint(2), max(0ll, N - i - 2));
	}
	mint ans = 0;
	REP(i, N + 1) {
		out(i, dp[i][0][2]);
		ans += dp[i][0][2];
	}
	out(ans / (modpow((mint)2, N - 1)));
}
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