#1008140 (C++17) No.2872 Depth of the Parentheses

• 問題: No.2872 Depth of the Parentheses
• ユーザー: tnakao0123
• 提出日時: 2024-09-07 14:49:35
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 1 ms / 2,000 ms
• コード長: 2,070 bytes
• コンパイル時間: 555 ms
• コンパイル使用メモリ: 43,392 KB
• 最終ジャッジ日時: 2025-02-24 05:08:42
• ジャッジサーバーID （参考情報）: judge4 / judge1

ファイルパターン	結果
sample	AC * 3
other	AC * 20 RE * 5

コンパイルメッセージ

main.cpp: In function ‘int main()’:
main.cpp:66:8: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   66 |   scanf("%d%d", &x, &k);
      |   ~~~~~^~~~~~~~~~~~~~~~

ソースコード

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/* -*- coding: utf-8 -*-
 *
 * 2872.cc:  No.2872 Depth of the Parentheses - yukicoder
 */
#include<cstdio>
#include<algorithm>
using namespace std;
/* constant */
const int MAX_K = 10;
const int MAX_K2 = MAX_K * 2;
const int MOD = 998244353;
/* 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 dp[MAX_K2 + 1][MAX_K + 1][MAX_K + 1];
/* subroutines */
/* main */
int main() {
  int x, k;
  scanf("%d%d", &x, &k);
  int k2 = k * 2;
  mi p0 = mi(x) / 100, p1 = mi(1) - p0;
  dp[0][0][0] = 1;
  for (int i = 0; i < k2; i++) // current pos
    for (int j = 0; j <= k; j++) // current depth
      for (int m = 0; m <= k; m++) // max depth
    if ((int)dp[i][j][m]) {
      if (j < k) { // put '(' in prob p0
        int m1 = max(j + 1, m);
        dp[i + 1][j + 1][m1] += dp[i][j][m] * p0;
      }
      if (j > 0) { // put ')' in prob p1
        int m1 = max(j, m);
        dp[i + 1][j - 1][m1] += dp[i][j][m] * p1;
      }
    }
  mi sum = 0;
  for (int m = 0; m <= k; m++) sum += dp[k2][0][m] * m;
  printf("%d\n", (int)sum);
  
  return 0;
}
 
 
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