Submission #1049593

#	Time	Username	Problem	Language	Result	Execution time	Memory
1049593		GrindMachine	Sparklers (JOI17_sparklers)	C++17	50 / 100	106 ms	262144 KiB

sparklers

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*

refs:
edi
stress tested to figure out the bug (ans = 0 case was bugged)
ac code used for stress testing: https://oj.uz/submission/1043916

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n,k,t; cin >> n >> k >> t;
    vector<ll> a(n+5);
    rep1(i,n) cin >> a[i];

    auto ok = [&](ll s){
        bool dp[n+5][n+5];
        memset(dp,0,sizeof dp);
        dp[k][k] = 1;
        rev(l,n,1){
            for(int r = l; r <= n; ++r){
                ll len = r-l+1;
                if(!dp[l][r]) conts;

                __int128 lim = (__int128)2*len*t*s;

                // l --> l-1
                if(l-1){
                    ll m = a[l-1]+a[r];
                    if(abs(m-2*a[l-1]) <= lim and abs(m-2*a[r]) <= lim){
                        dp[l-1][r] = 1;
                    }
                }

                // r --> r+1
                if(r+1 <= n){
                    ll m = a[l]+a[r+1];
                    if(abs(m-2*a[r+1]) <= lim and abs(m-2*a[l]) <= lim){
                        dp[l][r+1] = 1;
                    }
                }
            }
        }

        return dp[1][n];
    };

    ll lo = 0, hi = inf1;
    ll ans = -1;

    while(lo <= hi){
        ll mid = (lo+hi) >> 1;
        if(ok(mid)){
            ans = mid;
            hi = mid-1;
        }
        else{
            lo = mid+1;
        }
    }

    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 30 / 30
• Subtask 2: 20 / 20
• Subtask 3: 0 / 50