Submission #284381

#	Time	Username	Problem	Language	Result	Execution time	Memory
284381		AKaan37	Boxes with souvenirs (IOI15_boxes)	C++17	35 / 100	8 ms	388 KiB

boxes

//Bismillahirrahmanirrahim

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

using namespace std;

typedef long long lo;
typedef pair< lo,lo > PII;

#define fi first
#define se second
#define mp make_pair
#define endl "\n"
#define pb push_back
#define fio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define FOR for(int i=1;i<=n;i++)
#define mid ((start+end)/2)
#define ort ((bas+son)/2)

const lo inf = 1000000000000000000;
const lo KOK = 100000;
const lo LOG = 30;
const lo li = 500005;
const lo mod = 1000000007;

int n,m,b[li],a[li],k,flag,t;
lo cev;
vector<lo> v,vv;

long long delivery(int N, int K, int L, int p[]) {
	//~ cout<<"**\n";
	for(int i=0;i<N;i++){
		if(p[i]==0)continue;
		if(p[i]<=(L-p[i]))v.pb(p[i]);
		else vv.pb(p[i]);
	}
	n=N;
	k=K;
	lo mn=inf;
	sort(v.begin(),v.end());
	sort(vv.begin(),vv.end());
	reverse(v.begin(),v.end());
	int bas=0;
	int son=0;
	
	cev=0;
	for(int i=1;i<=n;i++){
		if(i%K==1%K)cev+=L;
		if(bas==(int)v.size())son++;
		else if(son==(int)vv.size())bas++;
		else{
			bas++;
			lo cev1=0;
			for(int j=bas;j<(int)v.size();j++){
				if(j%k==bas%k)cev1+=(v[j])*2;
			}
			for(int j=son;j<(int)vv.size();j++){
				if(j%k==son%k)cev1+=(L-vv[j])*2;
			}
			bas--;
			son++;
			lo cev2=0;
			for(int j=bas;j<(int)v.size();j++){
				if(j%k==bas%k)cev2+=(v[j])*2;
			}
			for(int j=son;j<(int)vv.size();j++){
				if(j%k==son%k)cev2+=(L-vv[j])*2;
			}
			son--;
			if(cev1<=cev2)bas++;
			else son++;
			mn=min(mn,cev1+cev);
			continue;
		}
		lo cev1=0;
		for(int j=bas;j<(int)v.size();j++){
			if(j%k==bas%k)cev1+=(v[j])*2;
		}
		for(int j=son;j<(int)vv.size();j++){
			if(j%k==son%k)cev1+=(L-vv[j])*2;
		}
		mn=min(mn,cev1+cev);
	}
	bas=0;
	son=0;
	lo cev1=0;
	for(int j=bas;j<(int)v.size();j++){
		if(j%k==bas%k)cev1+=(v[j])*2;
	}
	for(int j=son;j<(int)vv.size();j++){
		if(j%k==son%k)cev1+=(L-vv[j])*2;
	}
	//~ cout<<cev1<<"**\n"<<endl;
	mn=min(mn,cev1);
    return mn;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 15 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30