제출 #941579

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
941579		Xiaoyang	A Difficult(y) Choice (BOI21_books)	C++17	0 / 100	1 ms	476 KiB

books

#include "books.h"
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
 
#define fi first 
#define se second 
#define pll pair<ll,ll>
#define pb push_back
#define debug(x) cerr<<#x<<"="<<x<<endl;
#define MP make_pair
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define SZ(x) (ll)x.size()
#define ALL(x) x.begin(),x.end()
#define endl "\n"
const ll inf=1e18;
 
const ll maxn=1e5+5;
ll diff[maxn];
 
ll d(ll x){
	if(diff[x])return diff[x];
	return diff[x]=skim(x);
}
void solve(int n, int k, long long A, int s) {
	ll sum=0;
	vector<int>ans;
	rep(i,1,n+1){
		if(i<=k)sum+=d(i),ans.pb(i);
	}
	if(sum>2*A)impossible();
	if(sum>=A)answer(ans);
	// the sum of the first k elements is def lesser than A
	//if >=A and <=2A we would have alr answered
	//if >2A its def impossible 
	//if we find the largest index of element that is lesser than A

	ll lo=1,hi=n;
	while(lo<hi){
		ll mid=(lo+hi+1)>>1;
		if(d(mid)<=A)lo=mid;
		else hi=mid-1;
	}

	//lo+1 is more than A, check if >A + <A <=2A
	if(lo+1<=n and sum-d(k)+d(lo+1)<=2*A){
		ans[k-1]=lo;
		answer(ans);
	}
	
	//if not uhhh :think 
	//greedy? 
	//since lo is like less than A, and everything in the [1,k] is def<A
	//going from [1,k] to [2,k]U[lo] will def be <=2A
	//but if this continues we might get to >A eventually?
	//but as soon as its more than A we are done?
	//if it ever gets to >A then we just check its <=2A?
	//if we do liddis i dont think we will have chance to exceed 2A???
	ll id=0;
	while(id<k and lo>0){
		if(sum>A){
			answer(ans);
		}
		ans[id]=lo;
		sum-=d(id);
		sum+=d(lo);
		id++;
		lo--;
	}
	impossible();
	
	
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 15
• Subtask 3: 0 / 10
• Subtask 4: 0 / 15
• Subtask 5: 0 / 15
• Subtask 6: 0 / 20
• Subtask 7: 0 / 20