結果
| 問題 | No.1513 simple 門松列 problem |
| コンテスト | |
| ユーザー |
leaf_1415
|
| 提出日時 | 2021-05-21 22:09:19 |
| 言語 | C++11(廃止可能性あり) (gcc 13.3.0 + boost 1.89.0) |
| 結果 |
AC
|
| 実行時間 | 300 ms / 3,000 ms |
| コード長 | 6,218 bytes |
| 記録 | |
| コンパイル時間 | 1,080 ms |
| コンパイル使用メモリ | 88,812 KB |
| 実行使用メモリ | 138,664 KB |
| 最終ジャッジ日時 | 2024-10-10 08:52:45 |
| 合計ジャッジ時間 | 3,602 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 18 |
ソースコード
#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;
}
leaf_1415