Submission #1096925

#	Time	Username	Problem	Language	Result	Execution time	Memory
1096925		nguyen31hoang08minh2003	Rack (eJOI19_rack)	C++17	100 / 100	10 ms	11824 KiB

rack

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <math.h>
#include <bitset>
#include <vector>
#include <string>
#include <cstdio>
#include <cctype>
#include <numeric>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <iomanip>
#include <cassert>
#include <cstring>
#include <stdio.h>
#include <string.h>
#include <iterator>
#include <iostream>
#include <algorithm>
#include <strings.h>
#include <functional>
#include <unordered_set>
#include <unordered_map>
#define fore(i, a, b) for (int i = (a), i##_last = (b); i < i##_last; ++i)
#define fort(i, a, b) for (int i = (a), i##_last = (b); i <= i##_last; ++i)
#define ford(i, a, b) for (int i = (a), i##_last = (b); i >= i##_last; --i)
#define fi first
#define se second
#define pb push_back
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using namespace std;
using ll = long long;
using ld = long double;

template<class A, class B> bool maxi(A &a, const B &b) {return (a < b) ? (a = b, true):false;};
template<class A, class B> bool mini(A &a, const B &b) {return (a > b) ? (a = b, true):false;};

typedef unsigned long long ull;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;

constexpr int MAX_N = 1E6 + 6, MOD = 1E9 + 7;
const ll INF = 0X3F3F3F3F3F3F3F3F;

signed result, p[MAX_N];
ll k, r[MAX_N];
int n;

/*

    Idea:
        At a node, if the weights of two children branches are equal than
            go to left branch
        Otherwise
            go to right branch

    Observation:
        At i-th level
            there is power(2, i - 1) direct branches from the immediately above level

            and every branch of these branches will be visited before
                any branch is visited again

            This means that,
                there will be power(2, i - 1) times of choosing the left child branch
                then power(2, i - 1) times of choosing the right child branch
                then again ... (continue the cycle)

*/

signed main() {

    #ifdef LOCAL
    freopen("input.INP", "r", stdin);
//    freopen("output.OUT", "w", stdout);
    #endif // LOCAL
    cin.tie(0) -> sync_with_stdio(0);
    cout.tie(0);

    cin >> n >> k;

    p[0] = r[0] = 1;

    fort(i, 1, n) {
        p[i] = (p[i - 1] << 1) % MOD;
        r[i] = min(r[i - 1] << 1, INF);
    }

    for (int i = n - 1, j = 0; i >= 0; --i, ++j)
        if (((k - 1) / r[j]) & 1)
            (result += p[i]) %= MOD;

    (++result) %= MOD;

    cout << result << '\n';

    return 0;
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 60 / 60