Submission #371418

#	Time	Username	Problem	Language	Result	Execution time	Memory
371418		Sorting	Gap (APIO16_gap)	C++17	30 / 100	55 ms	1260 KiB

gap

#include "gap.h"
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
using namespace std;

void check_max(ll &a, ll b){ a = (b > a) ? b : a; }

const ll INF = 1e18;
const int INT_INF =  1e9;

int t, n;

ll solve_1(){
    ll l, r, ans = 0;
    MinMax(0, INF, &l, &r);

    ll l2, r2;
    n -= 2;
    while(n > 0){
        MinMax(l + 1, r - 1, &l2, &r2);
        ans = max(ans, max(l2 - l, r - r2));
        l = l2, r = r2;
        n -= 2;
    }
    ans = max(ans, r - l);
    return ans;
}

pair<ll, int> solve(ll l, ll r, int cnt, ll curr_mx){
    if(r - l <= curr_mx) return {0, -1};
    if(!cnt) return {0, 0};

    ll mid = (l + r) >> 1, l2, r2, l3, r3;
    MinMax(l + 1, mid, &l2, &r2);
    MinMax(mid + 1, r - 1, &l3, &r3);
    if(l2 == -1 && l3 == -1) return {r - l, 0};

    if(l2 != -1) curr_mx = max(l2 - l, curr_mx);
    else curr_mx = max(l3 - l, curr_mx);

    if(r3 != -1) curr_mx = max(r - r3, curr_mx);
    else curr_mx = max(r - r2, curr_mx);

    if(l2 != -1 && l3 != -1) curr_mx = max(l3 - r2, curr_mx);

    int new_cnt = 0;
    if(l2 != -1){
        auto p = solve(l2, r2, -1, curr_mx);
        check_max(curr_mx, p.first);
        if(p.second == -1){
            new_cnt -= INT_INF;
            cnt = -1;
        }

        if(cnt != -1) cnt -= p.second;
        else new_cnt += p.second;
    }
    if(l3 != -1){
        auto p = solve(l2, r2, cnt, curr_mx);
        check_max(curr_mx, p.first);
        if(p.second == -1){
            new_cnt -= INT_INF;
            cnt = -1;
        }

        if(cnt != -1) cnt -= p.second;
        else new_cnt += p.second;
    }

    if(cnt == -1) new_cnt = -1;
    if(new_cnt < 0) new_cnt = -1;
    return {curr_mx, new_cnt};
}

ll findGap(int _t, int _n){
	t = _t, n = _n;
    if(t == 1) return solve_1();
    if(n <= 1) return 0;
    ll l, r;
    MinMax(0, INF, &l, &r);
    ll curr_mx = ((r - l) / (n - 1)) + !!((r - l) % (n - 1));
    return max(solve(l, r, n - 2, curr_mx).first, curr_mx);
}

• Subtask 1: 30 / 30
• Subtask 2: 0 / 70