Submission #1200640

#	Time	Username	Problem	Language	Result	Execution time	Memory
1200640		countless	Hexagonal Territory (APIO21_hexagon)	C++20	9 / 100	0 ms	328 KiB

hexagon

#include "hexagon.h"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

const ll MOD = 1e9+7;
const ll INF = 1e18;
const ld EPS = 1e-12;

#define endl "\n"
#define sp <<" "<<
#define REP(i, a, b) for(ll i = a; i < b; i++)
#define dbg(x) cout << #x << " = " << x << endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fast_io() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

template <typename Key, typename Value>
using hash_map = unordered_map<Key, Value, custom_hash>;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// uniform_int_distribution<int>(a, b)(rng);
// shuffle(all(a), rng);

ll binpow(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) {
            res = (res * a) % MOD;
        }
        a = (a * a) % MOD;
        b >>= 1;
    }
    return res;
}

int draw_territory(int N, int A, int B, vector<int> D, vector<int> L) {
    assert(N == 3);
    assert(L[0] == L[1] and L[1] == L[2]);

    ll len = L[0]; len;
    // d elements with: A + (d-1) * B
    // calc separate terms
    int term1 = (1LL * ll(A) * (((len + 1) * (len + 2) / 2) % MOD)) % MOD;
    int term2 = (1LL * ll(B) * (((((((len * (len + 1)) % MOD) * (2LL * len + 1)) % MOD) * binpow(6, MOD - 2)) % MOD) + ((len * (len + 1) / 2) % MOD) % MOD)) % MOD;
    int ans = term1 + term2;
    ans %= MOD;
    return ans;
}

• Subtask 1: 3 / 3
• Subtask 2: 6 / 6
• Subtask 3: 0 / 11
• Subtask 4: 0 / 12
• Subtask 5: 0 / 15
• Subtask 6: 0 / 19
• Subtask 7: 0 / 18
• Subtask 8: 0 / 16