結果

問題 No.1549 [Cherry 2nd Tune] BANning Tuple
ユーザー leaf_1415leaf_1415
提出日時 2021-06-11 23:58:01
言語 C++11
(gcc 11.4.0)
結果
WA  
実行時間 -
コード長 7,907 bytes
コンパイル時間 2,796 ms
コンパイル使用メモリ 98,216 KB
実行使用メモリ 18,148 KB
最終ジャッジ日時 2024-05-08 20:40:44
合計ジャッジ時間 12,352 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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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
実行使用メモリ
testcase_00 AC 27 ms
6,816 KB
testcase_01 WA -
testcase_02 AC 3,723 ms
5,376 KB
testcase_03 TLE -
testcase_04 -- -
testcase_05 -- -
testcase_06 -- -
testcase_07 -- -
testcase_08 -- -
testcase_09 -- -
testcase_10 -- -
testcase_11 -- -
testcase_12 -- -
testcase_13 -- -
testcase_14 -- -
testcase_15 -- -
testcase_16 -- -
testcase_17 -- -
testcase_18 -- -
testcase_19 -- -
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ソースコード

diff # 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(llint (x) = (s); (x) >= (t); (x)--)
#define reps(x, s) for(llint (x) = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

//const ll mod = 1000000007;
const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}
	
	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "YES" << endl; }
void no(){ cout << "NO" << endl; }
ll sgn(ll x){ if(x > 0) return 1; if(x < 0) return -1; return 0;}
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll lcm(ll a, ll b){return a/gcd(a, b)*b;}
ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;}
ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;}
string lltos(ll x){string ret; for(;x;x/=10) ret += x % 10 + '0'; reverse(ret.begin(), ret.end()); return ret;}
ll stoll(string &s){ll ret = 0; for(auto c : s) ret *= 10, ret += c - '0'; return ret;}
template<typename T>
void uniq(T &vec){ sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());}

template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++; if(itr != map_var.end()) os << ","; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, multiset<T>& set_var) {
	for(typename multiset<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t){
	for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}
	cout << endl;
}
void dump_func() {cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}

llint modpow(llint a, llint n, llint mod)
{
	if(n == 0) return 1;
	if(n % 2){
		return ((a%mod) * (modpow(a, n-1, mod)%mod)) % mod;
	}
	else{
		return modpow((a*a)%mod, n/2, mod) % mod;
	}
}

int rev(int x, int n)
{
	int ret = 0;
	for(int i = 0; i < n; i++){
		ret <<= 1;
		ret |= (x>>i) & 1;
	}
	return ret;
}

//f[]とF[]は異なる実体を持たなければならない。rootには1の原始2^n乗根を渡す
void DFT(llint f[], llint F[], int n, llint mod, llint root)
{
	int N = 1<<n;
	for(int i = 0; i < N; i++) F[rev(i, n)] = f[i];
	
	llint a, b, x, z;
	for(int i = 1; i <= n; i++){
		int l = 1<<i;
		z = modpow(root, 1<<(n-i), mod);
		for(int j = 0; j < N/l; j++){
			x = 1;
			for(int k = 0; k < l/2; k++){
				a = F[j*l+k], b = F[j*l+k+l/2];
				F[j*l+k] = a + x * b % mod;
				F[j*l+k+l/2] = a - x * b % mod + mod;
				if(F[j*l+k] >= mod) F[j*l+k] -= mod;
				if(F[j*l+k+l/2] >= mod) F[j*l+k+l/2] -= mod;
				x *= z, x %= mod;
			}
		}
	}
}

//f[]とF[]は異なる実体を持たなければならない。rootには1の原始2^n乗根を渡す
void IDFT(llint F[], llint f[], int n, llint mod, llint root)
{
	int N = 1<<n;
	for(int i = 0; i < N; i++) f[rev(i, n)] = F[i];
	root = modpow(root, mod-2, mod);
	
	llint a, b, x, z;
	for(int i = 1; i <= n; i++){
		int l = 1<<i;
		z = modpow(root, 1<<(n-i), mod);
		for(int j = 0; j < N/l; j++){
			x = 1;
			for(int k = 0; k < l/2; k++){
				a = f[j*l+k], b = f[j*l+k+l/2];
				f[j*l+k] = a + x * b % mod;
				f[j*l+k+l/2] = a - x * b % mod + mod;
				if(f[j*l+k] >= mod) f[j*l+k] -= mod;
				if(f[j*l+k+l/2] >= mod) f[j*l+k+l/2] -= mod;
				x *= z, x %= mod;
			}
		}
	}
	llint Ninv = modpow(N, mod-2, mod);
	for(int i = 0; i < N; i++) f[i] *= Ninv, f[i] %= mod;
}

ll n, Q, nid;
ll f[105][1<<13], F[105][1<<13];
ll g[1<<13], G[1<<13];
vector<P> vec[105];
map<ll, ll> mp;
ll used[3005];

ll calc(ll t)
{
	ll ret = g[t], cb = 1;
	rep(i, 1, t){
		cb *= (n-nid+i)%mod, cb %= mod;
		cb *= modpow(i, mod-2, mod), cb %= mod;
		ret += g[t-i] * cb % mod;
		ret %= mod;
	}
	return ret;
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> Q;
	ll root = modpow(modpow(3, 119, mod), (1<<10), mod);
	ll N = 1<<13;
	
	nid = 0;
	ll k, a, b, s, t;
	rep(q, 1, Q){
		cin >> k >> a >> b >> s >> t;
		if(mp.count(k) == 0) mp[k] = ++nid;
		ll id = mp[k];
		vec[id].push_back(P(a, b));
		
		rep(i, 0, 3000) used[i] = 0;
		for(auto p : vec[id]){
			used[p.first]++, used[p.second+1]--;
		}
		rep(i, 1, 3000) used[i] += used[i-1];
		
		rep(i, 0, N-1) f[id][i] = 0;
		rep(i, 0, 3000) if(used[i] == 0) f[id][i] = 1;
		DFT(f[id], F[id], 13, mod, root);
		
		rep(i, 0, N-1) G[i] = 1, g[i] = 0;
		rep(i, 1, nid){
			rep(j, 0, N-1) G[j] *= F[i][j], G[j] %= mod;
			IDFT(G, g, 13, mod, root);
			rep(j, 3001, N-1) g[j] = 0;
			DFT(g, G, 13, mod, root);
		}
		ll ans = (calc(t) - calc(s-1) + mod) % mod;
		outl(ans);
	}
	
	return 0;
}
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