제출 #484719

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
484719		jalsol	말 (IOI15_horses)	C++17	17 / 100	1592 ms	14352 KiB

horses

#include "horses.h"
#ifdef LOCAL
#include "local.h"
#else
#include <bits/stdc++.h>
#define debug(...)
#define DB(...)
#endif

using namespace std;

const bool __initialization = []() {
    cin.tie(nullptr)->sync_with_stdio(false);
    cout << setprecision(12) << fixed;
    return true;
}();

using ll = long long;
using ld = long double;

#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

#define For(i, l, r) for (int i = (l); i <= (r); ++i)
#define Ford(i, r, l) for (int i = (r); i >= (l); --i)
#define Rep(i, n) For (i, 0, (n) - 1)
#define Repd(i, n) Ford (i, (n) - 1, 0)
#define Fe(...) for (auto __VA_ARGS__)

template <class C> inline int isz(const C& c) { return static_cast<int>(c.size()); }
template <class T> inline bool chmin(T& a, const T& b) { return (a > b) ? a = b, true : false; }
template <class T> inline bool chmax(T& a, const T& b) { return (a < b) ? a = b, true : false; }

constexpr ld eps = 1e-9;
constexpr int inf = 1e9;
constexpr ll linf = 1e18;

// =============================================================================

template <typename T> T mod_inv_in_range(T a, T m) {
    // assert(0 <= a && a < m);
    T x = a, y = m;
    T vx = 1, vy = 0;
    while (x) {
        T k = y / x;
        y %= x;
        vy -= k * vx;
        std::swap(x, y);
        std::swap(vx, vy);
    }
    assert(y == 1);
    return vy < 0 ? m + vy : vy;
}

template <typename T> T mod_inv(T a, T m) {
    a %= m;
    a = a < 0 ? a + m : a;
    return mod_inv_in_range(a, m);
}

template <int MOD_> struct modnum {
    static constexpr int MOD = MOD_;
    static_assert(MOD_ > 0, "MOD must be positive");

private:
    using ll = long long;

    int v;

public:

    modnum() : v(0) {}
    modnum(ll v_) : v(int(v_ % MOD)) { if (v < 0) v += MOD; }
    explicit operator int() const { return v; }
    friend std::ostream& operator << (std::ostream& out, const modnum& n) { return out << int(n); }
    friend std::istream& operator >> (std::istream& in, modnum& n) { ll v_; in >> v_; n = modnum(v_); return in; }

    friend bool operator == (const modnum& a, const modnum& b) { return a.v == b.v; }
    friend bool operator != (const modnum& a, const modnum& b) { return a.v != b.v; }

    modnum inv() const {
        modnum res;
        res.v = mod_inv_in_range(v, MOD);
        return res;
    }
    friend modnum inv(const modnum& m) { return m.inv(); }
    modnum neg() const {
        modnum res;
        res.v = v ? MOD-v : 0;
        return res;
    }
    friend modnum neg(const modnum& m) { return m.neg(); }

    modnum operator- () const {
        return neg();
    }
    modnum operator+ () const {
        return modnum(*this);
    }

    modnum& operator ++ () {
        v ++;
        if (v == MOD) v = 0;
        return *this;
    }
    modnum& operator -- () {
        if (v == 0) v = MOD;
        v --;
        return *this;
    }
    modnum& operator += (const modnum& o) {
        v -= MOD-o.v;
        v = (v < 0) ? v + MOD : v;
        return *this;
    }
    modnum& operator -= (const modnum& o) {
        v -= o.v;
        v = (v < 0) ? v + MOD : v;
        return *this;
    }
    modnum& operator *= (const modnum& o) {
        v = int(ll(v) * ll(o.v) % MOD);
        return *this;
    }
    modnum& operator /= (const modnum& o) {
        return *this *= o.inv();
    }

    friend modnum operator ++ (modnum& a, int) { modnum r = a; ++a; return r; }
    friend modnum operator -- (modnum& a, int) { modnum r = a; --a; return r; }
    friend modnum operator + (const modnum& a, const modnum& b) { return modnum(a) += b; }
    friend modnum operator - (const modnum& a, const modnum& b) { return modnum(a) -= b; }
    friend modnum operator * (const modnum& a, const modnum& b) { return modnum(a) *= b; }
    friend modnum operator / (const modnum& a, const modnum& b) { return modnum(a) /= b; }
};

template <typename T> T pow(T a, long long b) {
    assert(b >= 0);
    T r = 1; while (b) { if (b & 1) r *= a; b >>= 1; a *= a; } return r;
}

using mint = modnum<inf + 7>;

constexpr int maxn = 5e5 + 5;

int n;
ll pw[maxn];
ll price[maxn];

int getAns() {
    ll ans = 1;
    ll exp = 1;

    For (i, 1, n) {
        exp *= pw[i];
        chmax(ans, exp * price[i]);
    }

    return int(mint(ans));
}

int init(int _N, int _X[], int _Y[]) {
    n = _N;
    For (i, 1, n) {
        pw[i] = _X[i - 1];
        price[i] = _Y[i - 1];
    }

    return getAns();
}

int updateX(int pos, int val) { 
    pw[pos + 1] = val;
    return getAns();
}

int updateY(int pos, int val) {
    price[pos + 1] = val;
    return getAns();
}

// selling everything at one single day I suppose?
// if there's an answer of selling at both days,
// then keeping instead of selling at the first day,
// keeping the number of horses grow then sell at the second day
// will profit more because of the property of exponential functions
// ig?
// but the values will get very, very large, very quickly...
// if it passes the first subtask, then the logic could be correct

• Subtask 1: 17 / 17
• Subtask 2: 0 / 17
• Subtask 3: 0 / 20
• Subtask 4: 0 / 23
• Subtask 5: 0 / 23