#1007900 (C++17) No.2874 Gunegune Tree

• 問題: No.2874 Gunegune Tree
• ユーザー: shobonvip
• 提出日時: 2024-09-06 23:07:11
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 74 ms / 2,000 ms
• コード長: 1,757 bytes
• コンパイル時間: 4,168 ms
• コンパイル使用メモリ: 254,140 KB
• 最終ジャッジ日時: 2025-02-24 04:53:00
• ジャッジサーバーID （参考情報）: judge5 / judge5

ファイルパターン	結果
sample	AC * 3
other	AC * 30

ソースコード

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#include<bits/stdc++.h>
using namespace std;
//* ATCODER
#include<atcoder/all>
using namespace atcoder;
typedef modint998244353 mint;
//*/
/* BOOST MULTIPRECISION
#include<boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
//*/
typedef long long ll;
#define rep(i, s, n) for (int i = (int)(s); i < (int)(n); i++)
#define rrep(i, s, n) for (int i = (int)(n)-1; i >= (int)(s); i--)
template <typename T> bool chmin(T &a, const T &b) {
    if (a <= b) return false;
    a = b;
    return true;
}
template <typename T> bool chmax(T &a, const T &b) {
    if (a >= b) return false;
    a = b;
    return true;
}
template <typename T> T max(vector<T> &a){
    assert(!a.empty());
    T ret = a[0];
    for (int i=0; i<(int)a.size(); i++) chmax(ret, a[i]);
    return ret;
}
template <typename T> T min(vector<T> &a){
    assert(!a.empty());
    T ret = a[0];
    for (int i=0; i<(int)a.size(); i++) chmin(ret, a[i]);
    return ret;
}
template <typename T> T sum(vector<T> &a){
    T ret = 0;
    for (int i=0; i<(int)a.size(); i++) ret += a[i];
    return ret;
}
int main(){
    int n; cin >> n;
    // 確率, 現在の期待値
    vector dp1(n+1, vector<mint>(5));
    vector dp2(n+1, vector<mint>(5));
    vector<mint> inv(10);
    rep(i,1,10){
        inv[i] = mint(i).inv();
    }
    dp1[1][0] = inv[5];
    dp1[1][1] = inv[5];
    dp1[1][2] = inv[5];
    dp1[1][3] = inv[5];
    dp1[1][4] = inv[5];
    dp2[1][1] = inv[5];
    dp2[1][2] = inv[5];
    dp2[1][3] = inv[5];
    rep(i,1,n){
        rep(j,0,5){
            int v = 0;
            rep(k,0,5){
                if (abs(j-k) <= 1){
                    v += 1;
                }
            }
            rep(k,0,5){
                if (abs(j-k) <= 1){
                    if (1<=k && k<=3){
                        dp2[i+1][k] += dp1[i][j] * inv[v];
                    }
                    dp2[i+1][k] += dp2[i][j] * inv[v];
                    dp1[i+1][k] += dp1[i][j] * inv[v];
                }
            }
        }
    }
    cout << sum(dp2[n]).val() << '\n';
}
 
 
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