結果

問題 No.1186 長方形の敷き詰め
ユーザー 👑 はまやんはまやんはまやんはまやん
提出日時 2020-08-22 15:12:08
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 27 ms / 2,000 ms
コード長 4,147 bytes
コンパイル時間 2,135 ms
コンパイル使用メモリ 204,428 KB
実行使用メモリ 18,792 KB
最終ジャッジ日時 2023-08-05 12:24:45
合計ジャッジ時間 4,013 ms
ジャッジサーバーID
(参考情報)
judge11 / judge15
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 27 ms
18,596 KB
testcase_01 AC 25 ms
18,788 KB
testcase_02 AC 25 ms
18,724 KB
testcase_03 AC 26 ms
18,592 KB
testcase_04 AC 24 ms
18,748 KB
testcase_05 AC 26 ms
18,468 KB
testcase_06 AC 26 ms
18,656 KB
testcase_07 AC 26 ms
18,664 KB
testcase_08 AC 27 ms
18,764 KB
testcase_09 AC 25 ms
18,724 KB
testcase_10 AC 27 ms
18,784 KB
testcase_11 AC 26 ms
18,568 KB
testcase_12 AC 25 ms
18,656 KB
testcase_13 AC 25 ms
18,596 KB
testcase_14 AC 25 ms
18,596 KB
testcase_15 AC 25 ms
18,776 KB
testcase_16 AC 25 ms
18,736 KB
testcase_17 AC 25 ms
18,732 KB
testcase_18 AC 25 ms
18,596 KB
testcase_19 AC 25 ms
18,512 KB
testcase_20 AC 26 ms
18,568 KB
testcase_21 AC 27 ms
18,572 KB
testcase_22 AC 27 ms
18,652 KB
testcase_23 AC 25 ms
18,568 KB
testcase_24 AC 25 ms
18,792 KB
testcase_25 AC 25 ms
18,668 KB
testcase_26 AC 26 ms
18,680 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<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
template<int MOD> struct ModInt {
    static const int Mod = MOD; unsigned x; ModInt() : x(0) { }
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const { long long a = x, b = MOD, u = 1, v = 0;
        while (b) { long long t = a / b; a -= t * b; std::swap(a, b); u -= t * v; std::swap(u, v); }
        return ModInt(u); }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ostream& operator<<(ostream& st, const ModInt<MOD> a) { st << a.get(); return st; };
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }
template<typename T, int FAC_MAX> struct Comb { vector<T> fac, ifac;
    Comb(){fac.resize(FAC_MAX,1);ifac.resize(FAC_MAX,1);rep(i,1,FAC_MAX)fac[i]=fac[i-1]*i;
        ifac[FAC_MAX-1]=T(1)/fac[FAC_MAX-1];rrep(i,FAC_MAX-2,1)ifac[i]=ifac[i+1]*T(i+1);}
    T aPb(int a, int b) { if (b < 0 || a < b) return T(0); return fac[a] * ifac[a - b]; }
    T aCb(int a, int b) { if (b < 0 || a < b) return T(0); return fac[a] * ifac[a - b] * ifac[b]; }
    // nHk = (n+k-1)Ck : n is separator (k balls to n boxs)
    T nHk(int n, int k) { if (n == 0 && k == 0) return T(1); if (n <= 0 || k < 0) return 0;
        return aCb(n + k - 1, k); }
    T pairCombination(int n) {if(n%2==1)return T(0);return fac[n]*ifac[n/2]/(T(2)^(n/2));}
    // combination of paris for n
}; 
typedef ModInt<998244353> mint;
/*---------------------------------------------------------------------------------------------------
            ∧_∧
      ∧_∧  (´<_` )  Welcome to My Coding Space!
     ( ´_ゝ`) /  ⌒i     @hamayanhamayan0
    /   \     | |
    /   / ̄ ̄ ̄ ̄/  |
  __(__ニつ/     _/ .| .|____
     \/____/ (u ⊃
---------------------------------------------------------------------------------------------------*/














int N, M;
Comb<mint, 2010101> com;
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> N >> M;

    if (M < N || N == 1) {
        cout << 1 << endl;
        return;
    }

    mint ans = 0;
    rep(yoko, 0, M / N + 1) {
        int rest = M - yoko * N;
        ans += com.aCb(rest + yoko, yoko);
    }
    cout << ans << endl;
}





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