제출 #1303589

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1303589		CSQ31	비스킷 담기 (IOI20_biscuits)	C++20	0 / 100	68 ms	63776 KiB

biscuits

#include "biscuits.h"
#include  <bits/stdc++.h>
using namespace std;

#define rep(i, a, b) for(int i=a; i<(b); ++i)
#define all(a) begin(a), end(a)
#define sz(x) (int)(x).size()
#define fi first
#define se second
#define pb push_back

template<typename T> using v = vector<T>;
template<typename T> using vv = v<vector<T>>;

typedef long long ll;
typedef pair<ll ,ll > pii;
typedef v<int> vi;
//Represent y as a subset of indices
//yi = {0,x}
//a0 >= y0
//a1 + a0/2 >= y1 + y0/2
//a2 + a1/2 + a0 >= y2 + y1/2 + y0/4
//...
//mutiply the ith eqn by 2^i, we get
//a0 >= y0
//2a1 + a0 >= 2y1 + y0
//4a2 + 2a1 + a0 >= 4y2 + 2y1 + y0
//2^k*ak + .... + a0 >= 2^k*yk + ... y0

//Let S = set of possible values of y0+2y1+4y2 + ... 2^i * yi 
//after processing a0 to ai

//we need to do two things
//1) S = S U 2^i * x
//2) truncate S by p(i) = sum_j<=i 2^j * aj
struct node{
	ll sum = 0;
	node * L = NULL;
	node * R = NULL;
	node(){}
	node(node *v){
		L = v->L;
		R = v->R;
		sum = v->sum;
	}
};
node* del(node *v,ll k,ll lvl){
	assert(lvl >= 0);
	node *u = new node(v);
	if(k == (1LL<<lvl))return u; //keep everything
	if(k == 0)return NULL; //destroy everything

	//cout<<k<<" "<<lvl<<" "<<u->sum<<'\n';
	if(k > (1LL<<(lvl-1))){
		if(u->R)u->R = del(u->R,k-(1LL<<(lvl-1)),lvl-1);
	}else{
		u->R = NULL;
		if(u->L)u->L = del(u->L,k,lvl-1);
	}
	u->sum = 0;
	if(u->L)u->sum += u->L->sum;
	if(u->R)u->sum += u->R->sum;
	//cout<<k<<" "<<lvl<<" "<<u->sum<<'\n';
	return u;
}

const ll INF = 2e18;
ll count_tastiness(ll x, vector<ll> a) {
	ll k = sz(a);
	vector<ll>p(60,0);
	for(int i=0;i<k;i++)p[i] = a[i]* (1LL<<i);
	for(int i=1;i<60;i++)p[i] += p[i-1];
	for(ll &v:p)v/=x;
	//for(ll v:p)cout<<v<<" ";
	//cout<<'\n';
	node *root = new node();
	root->sum = 1;
	for(ll i=0;i<60;i++){
		node *nroot = new node();
		nroot->L = new node(root);
		nroot->R = new node(root);
		nroot->sum = 2*root->sum;
		

		root = nroot;
		root = del(root,p[i]+1,i+1);
	}
	return root->sum;
}

• Subtask 1: 0 / 9
• Subtask 2: 0 / 12
• Subtask 3: 0 / 21
• Subtask 4: 0 / 35
• Subtask 5: 0 / 23