結果

問題 No.1513 simple 門松列 problem
ユーザー leaf_1415leaf_1415
提出日時 2021-05-21 22:09:19
言語 C++11
(gcc 11.4.0)
結果
AC  
実行時間 292 ms / 3,000 ms
コード長 6,218 bytes
コンパイル時間 887 ms
コンパイル使用メモリ 88,112 KB
実行使用メモリ 138,668 KB
最終ジャッジ日時 2024-04-18 15:38:51
合計ジャッジ時間 3,731 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ（β）

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 36 ms
138,572 KB
testcase_01 AC 36 ms
138,520 KB
testcase_02 AC 282 ms
138,448 KB
testcase_03 AC 36 ms
138,524 KB
testcase_04 AC 35 ms
138,652 KB
testcase_05 AC 36 ms
138,652 KB
testcase_06 AC 37 ms
138,648 KB
testcase_07 AC 37 ms
138,648 KB
testcase_08 AC 38 ms
138,652 KB
testcase_09 AC 36 ms
138,668 KB
testcase_10 AC 35 ms
138,528 KB
testcase_11 AC 35 ms
138,652 KB
testcase_12 AC 37 ms
138,528 KB
testcase_13 AC 36 ms
138,652 KB
testcase_14 AC 292 ms
138,648 KB
testcase_15 AC 288 ms
138,640 KB
testcase_16 AC 278 ms
138,584 KB
testcase_17 AC 193 ms
138,656 KB
testcase_18 AC 85 ms
138,656 KB
testcase_19 AC 39 ms
138,652 KB
testcase_20 AC 66 ms
138,604 KB
権限があれば一括ダウンロードができます

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)...);
}

ll n, m;
mint dp[205][205][205], dp2[205][205][205];
mint sum[205][205], sum2[205][205];

mint get(mint sum[205][205], ll i, ll l, ll r)
{
	if(l > r) return mint(0);
	mint ret = sum[i][r];
	if(l > 0) ret -= sum[i][l-1];
	return ret;
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m;
	
	rep(j, 0, m-1) rep(k, 0, m-1)
	if(j != k){
		dp[2][j][k] = mint(1);
		dp2[2][j][k] = mint(j+k);
	}
	
	rep(i, 3, n){
		rep(j, 0, m-1){
			sum[j][0] = dp[i-1][j][0], sum2[j][0] = dp2[i-1][j][0];
			rep(k, 1, m-1){
				sum[j][k] = sum[j][k-1] + dp[i-1][j][k];
				sum2[j][k] = sum2[j][k-1] + dp2[i-1][j][k];
			}
		}
		rep(j, 0, m-1){
			rep(k, 0, m-1){
				if(j < k){
					dp[i][j][k] = get(sum, k, 0, j-1) + get(sum, k, j+1, k-1);
					dp2[i][j][k] = get(sum2, k, 0, j-1) + get(sum2, k, j+1, k-1) + dp[i][j][k] * mint(j);
				}
				if(j > k){
					dp[i][j][k] = get(sum, k, k+1, j-1) + get(sum, k, j+1, m-1);
					dp2[i][j][k] = get(sum2, k, k+1, j-1) + get(sum2, k, j+1, m-1) + dp[i][j][k] * mint(j);
				}
				//outl(i, j, k, dp[i][j][k]);
			}
		}
	}
	
	mint ans = 0, ans2 = 0;
	rep(j, 0, m-1) rep(k, 0, m-1) ans += dp[n][j][k], ans2 += dp2[n][j][k];
	outl(ans, ans2);
	
	return 0;
}
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