#1008792 (C++17) No.2874 Gunegune Tree

• 問題: No.2874 Gunegune Tree
• ユーザー: tnakao0123
• 提出日時: 2024-09-09 14:05:36
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 9 ms / 2,000 ms
• コード長: 2,381 bytes
• コンパイル時間: 646 ms
• コンパイル使用メモリ: 44,288 KB
• 最終ジャッジ日時: 2025-02-24 06:19:00
• ジャッジサーバーID （参考情報）: judge2 / judge5

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

コンパイルメッセージ

main.cpp: In function ‘int main()’:
main.cpp:68:8: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   68 |   scanf("%d", &n);
      |   ~~~~~^~~~~~~~~~

ソースコード

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/* -*- coding: utf-8 -*-
 *
 * 2874.cc:  No.2874 Gunegune Tree - yukicoder
 */
#include<cstdio>
#include<algorithm>
using namespace std;
/* constant */
const int MAX_N = 200000;
const int K = 5;
const int MOD = 998244353;
enum { R, RU, U, LU, L };
/* typedef */
template<const int MOD>
struct MI {
  int v;
  MI(): v() {}
  MI(int _v): v(_v % MOD) { if (v < 0) v += MOD; }
  MI(long long _v): v(_v % MOD) { if (v < 0) v += MOD; }
  explicit operator int() const { return v; }
  
  MI operator+(const MI m) const { return MI(v + m.v); }
  MI operator-(const MI m) const { return MI(v + MOD - m.v); }
  MI operator*(const MI m) const { return MI((long long)v * m.v); }
  MI &operator+=(const MI m) { return (*this = *this + m); }
  MI &operator-=(const MI m) { return (*this = *this - m); }
  MI &operator*=(const MI m) { return (*this = *this * m); }
  bool operator==(const MI m) const { return v == m.v; }
  bool operator!=(const MI m) const { return v != m.v; }
  MI pow(int n) const {  // a^n % MOD
    MI pm = 1, a = *this;
    while (n > 0) {
      if (n & 1) pm *= a;
      a *= a;
      n >>= 1;
    }
    return pm;
  }
  MI inv() const { return pow(MOD - 2); }
  MI operator/(const MI m) const { return *this * m.inv(); }
  MI &operator/=(const MI m) { return (*this = *this / m); }
};
using mi = MI<MOD>;
/* global variables */
mi invs[K + 1], dp[MAX_N + 1][K];
/* subroutines */
/* main */
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= K; i++) invs[i] = mi(i).inv();
  for (int j = 0; j < K; j++) dp[1][j] = invs[K];
  for (int i = 1; i < n; i++) {
    // R -> R, RU
    dp[i + 1][R]  += dp[i][R] * invs[2];
    dp[i + 1][RU] += dp[i][R] * invs[2];
    // RU -> R, RU, U
    dp[i + 1][R]  += dp[i][RU] * invs[3];
    dp[i + 1][RU] += dp[i][RU] * invs[3];
    dp[i + 1][U]  += dp[i][RU] * invs[3];
    // U -> RU, U, LU
    dp[i + 1][RU] += dp[i][U] * invs[3];
    dp[i + 1][U]  += dp[i][U] * invs[3];
    dp[i + 1][LU] += dp[i][U] * invs[3];
    // LU -> L, LU, U
    dp[i + 1][L]  += dp[i][LU] * invs[3];
    dp[i + 1][LU] += dp[i][LU] * invs[3];
    dp[i + 1][U]  += dp[i][LU] * invs[3];
    // L -> L, LU
    dp[i + 1][L]  += dp[i][L] * invs[2];
    dp[i + 1][LU] += dp[i][L] * invs[2];
  }
  mi sum = 0;
  for (int i = 1; i <= n; i++)
    sum += dp[i][RU] + dp[i][U] + dp[i][LU];
  printf("%d\n", (int)sum);
  
  return 0;
}
 
 
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