結果

問題 No.802 だいたい等差数列
ユーザー momoharamomohara
提出日時 2024-04-13 14:44:03
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 66 ms / 2,000 ms
コード長 5,855 bytes
コンパイル時間 4,884 ms
コンパイル使用メモリ 320,744 KB
実行使用メモリ 26,336 KB
最終ジャッジ日時 2024-10-03 00:08:19
合計ジャッジ時間 7,011 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 30
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 <atcoder/all>
#include <bits/stdc++.h>

using namespace std;
using namespace atcoder;

using ll = long long;
using ull = unsigned long long;
using ld = long double;
using P = pair<ll, ll>;
using tp = tuple<ll, ll, ll>;

template <class T>
using vec = vector<T>;
template <class T>
using vvec = vector<vec<T>>;

#define all(hoge) (hoge).begin(), (hoge).end()
#define en '\n'
#define rep(i, m, n) for(ll i = (ll)(m); i < (ll)(n); ++i)
#define rep2(i, m, n) for(ll i = (ll)(n)-1; i >= (ll)(m); --i)
#define REP(i, n) rep(i, 0, n)
#define REP2(i, n) rep2(i, 0, n)

constexpr long long INF = 1LL << 60;
constexpr int INF_INT = 1 << 25;
constexpr long long MOD = (ll)1e9 + 7;
// constexpr long long MOD = 998244353LL;
static const ld pi = 3.141592653589793L;

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

template <class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

template <class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

struct Edge {
    int to, rev;
    ll cap;
    Edge(int _to, int _rev, ll _cap) : to(_to), rev(_rev), cap(_cap) {}
};

typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

void add_edge(Graph &G, int from, int to, ll cap, bool revFlag, ll revCap) {
    G[from].push_back(Edge(to, (int)G[to].size(), cap));
    if(revFlag)
        G[to].push_back(Edge(from, (int)G[from].size() - 1, revCap));
}

template <int mod>
struct ModInt {
    int x;

    ModInt() : x(0) {}

    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    ModInt &operator+=(const ModInt &p) {
        if((x += p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator-=(const ModInt &p) {
        if((x += mod - p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator*=(const ModInt &p) {
        x = (int)(1LL * x * p.x % mod);
        return *this;
    }

    ModInt &operator/=(const ModInt &p) {
        *this *= p.inverse();
        return *this;
    }

    ModInt operator-() const { return ModInt(-x); }

    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

    bool operator==(const ModInt &p) const { return x == p.x; }

    bool operator!=(const ModInt &p) const { return x != p.x; }

    ModInt inverse() const {
        int a = x, b = mod, u = 1, v = 0, t;
        while(b > 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
        }
        return ModInt(u);
    }

    ModInt pow(int64_t n) const {
        ModInt ret(1), mul(x);
        while(n > 0) {
            if(n & 1)
                ret *= mul;
            mul *= mul;
            n >>= 1;
        }
        return ret;
    }

    friend ostream &operator<<(ostream &os, const ModInt &p) {
        return os << p.x;
    }

    friend istream &operator>>(istream &is, ModInt &a) {
        int64_t t;
        is >> t;
        a = ModInt<mod>(t);
        return (is);
    }

    static int get_mod() { return mod; }
};

using mint = ModInt<MOD>;

template <class T>
struct Combination {
    vector<T> fact_, inv_, finv_;
    constexpr Combination() {}
    constexpr Combination(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
        init(n + 1);
    }
    constexpr void init(int n) noexcept {
        fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
        int MOD = fact_[0].get_mod();
        for(int i = 2; i < n; i++) {
            fact_[i] = fact_[i - 1] * i;
            inv_[i] = -inv_[MOD % i] * (MOD / i);
            finv_[i] = finv_[i - 1] * inv_[i];
        }
    }
    constexpr T nPr(int n, int k) const noexcept {
        if(n < k || n < 0 || k < 0)
            return 0;
        return fact_[n] * finv_[n - k];
    }
    constexpr T nCr(int n, int k) const noexcept {
        if(n < k || n < 0 || k < 0)
            return 0;
        if(n < MOD)
            return fact_[n] * finv_[k] * finv_[n - k];

        // Lucasの定理
        T ret = 1;
        while(n || k) {
            ll _n = n % MOD, _k = k % MOD;
            n /= MOD;
            k /= MOD;
            ret *= nCr(_n, _k);
        }
        return ret;
    }
    constexpr T nHr(int n, int k) const noexcept {
        if(n <= 0 || k < 0)
            return 0;
        return nCr(n + k - 1, k);
    }
    constexpr T fact(int n) const noexcept {
        if(n < 0)
            return 0;
        return fact_[n]; // n!
    }
    constexpr T inv(int n) const noexcept {
        if(n < 0)
            return 0;
        return inv_[n]; // 1/n
    }
    constexpr T finv(int n) const noexcept {
        if(n < 0)
            return 0;
        return finv_[n]; // 1/n!
    }
};

void solve() {
    ll n, m, d1, d2;
    cin >> n >> m >> d1 >> d2;

    Combination<mint> com(n + m);

    // f=(x^D1-x^(D2+1))^(N-1)
    vec<mint> f(m + 1, 0);
    REP(i, n) {
        ll k = d1 * (n - 1 - i) + (d2 + 1) * i;
        if(k > m)
            continue;
        if(i % 2)
            f[k] -= com.nCr(n - 1, i);
        else
            f[k] += com.nCr(n - 1, i);
    }

    // g=(1-x)^(-N-1)
    vec<mint> g(m + 1, 0);
    REP(i, m + 1) {
        g[i] = com.nCr(n + i, n);
    }

    mint ans = 0;
    REP(i, m) {
        ll j = m + 1 - i - 2;
        ans += f[i] * g[j];
    }
    cout << ans << en;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    // cout << fixed << setprecision(10);

    // ll t;
    // cin >> t;
    // REP(i, t - 1) {
    //     solve();
    // }

    solve();

    return 0;
}
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