結果

問題 No.916 Encounter On A Tree
コンテスト
ユーザー a
提出日時 2019-10-26 00:22:05
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,483 bytes
コンパイル時間 1,533 ms
コンパイル使用メモリ 169,052 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-09-13 11:20:29
合計ジャッジ時間 3,238 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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ファイルパターン 結果
sample AC * 4 WA * 1
other AC * 41 WA * 15
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 # 

//#define _GLIBCXX_DEBUG
#include "bits/stdc++.h"

using namespace std;

//------------------------------- Libraries --------------------------------//

template <int p>
struct Modint
{
    int value;

    Modint() : value(0) {}
    Modint(long x) : value(x >= 0 ? x % p : (p + x % p) % p) {}

    inline Modint &operator+=(const Modint &b)
    {
        if ((this->value += b.value) >= p)
            this->value -= p;
        return (*this);
    }
    inline Modint &operator-=(const Modint &b)
    {
        if ((this->value += p - b.value) >= p)
            this->value -= p;
        return (*this);
    }
    inline Modint &operator*=(const Modint &b)
    {
        this->value = (int)((1LL * this->value * b.value) % p);
        return (*this);
    }
    inline Modint &operator/=(const Modint &b)
    {
        (*this) *= b.inverse();
        return (*this);
    }

    Modint operator+(const Modint &b) const { return Modint(*this) += b; }
    Modint operator-(const Modint &b) const { return Modint(*this) -= b; }
    Modint operator*(const Modint &b) const { return Modint(*this) *= b; }
    Modint operator/(const Modint &b) const { return Modint(*this) /= b; }

    inline Modint &operator++(int) { return (*this) += 1; }
    inline Modint &operator--(int) { return (*this) -= 1; }

    inline bool operator==(const Modint &b) const
    {
        return this->value == b.value;
    }
    inline bool operator!=(const Modint &b) const
    {
        return this->value != b.value;
    }
    inline bool operator<(const Modint &b) const
    {
        return this->value < b.value;
    }
    inline bool operator<=(const Modint &b) const
    {
        return this->value <= b.value;
    }
    inline bool operator>(const Modint &b) const
    {
        return this->value > b.value;
    }
    inline bool operator>=(const Modint &b) const
    {
        return this->value >= b.value;
    }

    // requires that "this->value and p are co-prime"
    // a_i * v + a_(i+1) * p = r_i
    // r_i = r_(i+1) * q_(i+1) * r_(i+2)
    // q == 1 (i > 1)
    // reference: https://atcoder.jp/contests/agc026/submissions/2845729
    // (line:93)
    inline Modint inverse() const
    {
        assert(this->value != 0);
        int r0 = p, r1 = this->value, a0 = 0, a1 = 1;
        while (r1)
        {
            int q = r0 / r1;
            r0 -= q * r1;
            swap(r0, r1);
            a0 -= q * a1;
            swap(a0, a1);
        }
        return Modint(a0);
    }

    friend istream &operator>>(istream &is, Modint<p> &a)
    {
        long t;
        is >> t;
        a = Modint<p>(t);
        return is;
    }
    friend ostream &operator<<(ostream &os, const Modint<p> &a)
    {
        return os << a.value;
    }
};

const int MOD = 1e9 + 7;

using Int = Modint<MOD>;

//------------------------------- Type Names -------------------------------//

using i64 = int_fast64_t;

using seika = string;
//akari : 1D, yukari : 2D, maki : 3D vector
template <class kizuna>
using akari = vector<kizuna>;
template <class yuzuki>
using yukari = akari<akari<yuzuki>>;
template <class tsurumaki>
using maki = akari<yukari<tsurumaki>>;
//akane : ascending order, aoi : decending order
template <class kotonoha>
using akane = priority_queue<kotonoha, akari<kotonoha>, greater<kotonoha>>;
template <class kotonoha>
using aoi = priority_queue<kotonoha>;

//------------------------------- Dubug Functions ---------------------------//
inline void print()
{
    cout << endl;
}
template <typename First, typename... Rest>
void print(const First &first, const Rest &... rest)
{
    cout << first << ' ';
    print(rest...);
}
//------------------------------- Solver ------------------------------------//

void solve()
{
    int d, l, r, k;
    cin >> d >> l >> r >> k;
    l = log(l) / log(2) + 1;
    r = log(r) / log(2) + 1;
    int p = 1;
    //print(l, r);
    while (l + r - 2 * p > k)
    {
        p++;
    }
    if (l + r - 2 * p != k)
    {
        cout << 0 << endl;
        return;
    }
    l -= p - 1;
    r -= p - 1;
    //print(p, l, r);
    Int ans = 1;
    for (int x = 1; x <= d; x++)
    {
        int c = 1 << (x - 1);
        if (l + p - 1 == x)
            c--;
        if (r + p - 1 == x)
            c--;
        for (int y = 1; y <= c; y++)
            ans *= Int(y);
    }
    ans *= max(1, (1 << (r - 2)));
    ans *= max(1, (1 << (l - 2)));
    ans *= max(1, (1 << (p - 1 + (l != 1))));
    cout << ans << endl;
}

int main()
{
    solve();
    return 0;
}
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