제출 #462523

#제출 시각아이디문제언어결과실행 시간메모리
462523gesghaRack (eJOI19_rack)C++17
40 / 100
89 ms16724 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define fr(x, l, r) for(int x = l; x <= r; x++)
#define rf(x, l, r) for(int x = l; x >= r; x--)
#define fe(x, y) for(auto& x : y)

#define fi first
#define se second
#define m_p make_pair
#define pb push_back
#define pw(x) (ull(1) << ull(x))
#define all(x) (x).begin(),x.end()
#define sz(x) (int)x.size()
#define mt make_tuple
#define ve vector

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <ll,ll> pll;
typedef pair <int,int> pii;
typedef pair <ld,ld> pld;

template<typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key

template<typename T>
bool umn(T& a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T>
bool umx(T& a, T b) { return a < b ? a = b, 1 : 0; }

const ll inf = 1e18;
const int intf = 1e9;
const ll mod = 1e9 + 7;
const ld eps = 0.00000001;
const ll N  = 2e6;

ll t[4*N], n;


void modify(int v, int len){

    if(len == n){
        t[v] = 1;
        return;
    }

    if(t[v * 2 + 1] >= t[v*2]){
        modify(v * 2, len + 1);
    }
    else{
        modify(v * 2 + 1, len + 1);
    }
    t[v] = t[v * 2] + t[v * 2 + 1];
}


void get(ll& ans, ll v, ll len){

    if(len == n){
        ans = v;
        return;
    }

    if(t[v * 2 + 1] >= t[v*2]){
        get(ans,v * 2, len + 1);
    }
    else{
        get(ans, v * 2 + 1, len + 1);
    }
}

int main(){
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt","w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
#else
//    freopen("mountains.in", "r", stdin);
//    freopen("mountains.out","w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
#endif

    ll k;
    cin >> n >> k;

    while(--k){
        modify(1,0);
    }
    ll ans = 0;
//    fr(i,1, 5*(n)){
//        cout << i << " " << t[i] << endl;
//    }

    get(ans,1,0);
    cout << ans - pw(n) + 1 << endl;

    return 0;
}
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