Submission #1185007

#	Time	Username	Problem	Language	Result	Execution time	Memory
1185007		al95ireyiz	Rice Hub (IOI11_ricehub)	C++20	42 / 100	2 ms	1096 KiB

ricehub

//*** Bismillah ***//
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#if !defined(ONLINE_JUDGE) and !defined(EVAL)
#include "template/debug.h"
#else
#define d(x...)
#endif
#define fr first
#define sc second
#define in insert
#define ll long long
#define pb push_back
#define vll vector<ll>
#define pll pair<ll,ll>
#define all(x)x.begin(),x.end()
const ll INF = 1e9;
const ll INFL = 1e18;
const ll MOD = 1e9+7;
// const ll MOD = 998244353;
const ll maxn = 1e5+5;
ll n,m,k=0;
#include "ricehub.h"
ll a[maxn];
ll f(ll x, ll u){
    // x denesini min qiymete goturmeyin en yaxsi yolu en yaxin x denesinin medianini goturmekdi
    ll cv = INF, md = x/2 + u;
    // ilki elimle hesablayiram
    ll cm = 0;
    for(ll i=1;i<=x;i++){
        cm += abs(a[i] - a[md]);
    }
    cv = min(cv, cm);
    ll l=1, r=x, i = md, lsay = md - 1, rsay = r - md;
    while(r < n){
        l ++, r ++, i ++;
        cm -= abs(a[l-1] - a[i-1]);
        cm += abs(a[r] - a[i]);
        cm += abs(a[i-1] - a[i]) * lsay;
        cm -= abs(a[i-1] - a[i]) * rsay;
        cv = min(cv, cm);
    }
    return cv;
}
int besthub(int _R, int _L, int _X[], ll _B){
    n = _R, m = _L, k = _B;
    // memset(a, INF, sizeof(a));
    for(ll i=0;i<maxn;i++) a[i] = INFL;
    for(ll i=1;i<=n;i++) a[i] = _X[i-1];

    // Cavab - duyu zeminleri ucun binary search:
    int l = 0, r = n, cv=0;
    while(l <= r){
        ll md = (l + r) >> 1;
        if(min(f(md, 0), f(md, 1)) <= k) l = md + 1, cv = md;
        else r = md - 1;
    }
    return cv;
}
// void _(){
//     ll x, b, c;
//     cin>>x>>b>>c;
//     int d[x];
//     for(ll i=0;i<x;i++) cin>>d[i];
//     cout<<besthub(x, b, d, c)<<'\n';
// }
// signed main(){
//     ll tm=clock();
//     cin.tie(0)->sync_with_stdio(0);
//     ll t=1;
//     // cin>>t;
//     while(t--) _();
//     cerr<<"\n\033[1;31mTime: \033[1;30m"
//         <<(double)(clock()-tm)/1000000<<"\033[1;32m seconds\n";
// }

• Subtask 1: 17 / 17
• Subtask 2: 25 / 25
• Subtask 3: 0 / 26
• Subtask 4: 0 / 32