Submission #416762

#	Time	Username	Problem	Language	Result	Execution time	Memory
416762		jainbot27	A Difficult(y) Choice (BOI21_books)	C++17	0 / 100	2 ms	968 KiB

books

// Difficult(y) Choice

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

using ll=long long;

#define FOR(i, a, b) for(int i=a; i<b; i++) 
#define ROF(i, a, b) for(int i=b-1; i>=a; i--)
#define F0R(i, n) FOR(i, 0, n) 
#define R0F(i, n) ROF(i, 0, n) 

#define f first
#define s second
#define pb push_back
#define siz(x) (int)x.size() 

const int mxN=1e5+10;

int n, k, s; ll a;
ll x[mxN];

ll qry(int V){
    if(x[V]!=-1) return x[V]; 
    x[V]=skim(V+1);
    return x[V];
}

void chk(int X){
    if(X<0) return;
    if(X+k>=n) return;
    ll sum=0;
    vector<int> Ans;
    FOR(i, X, X+k){
        sum+=qry(i); 
        Ans.pb(i+1);
    }
    if(sum>=a&&sum<=2*a) answer(Ans); 
}
void solve(int N, int K, ll A, int S){
    n=N, k=K, a=A, s=S; 
    memset(x, -1, sizeof x);
    {
        ll sum=0; 
        F0R(i, k-1) sum+=qry(i);
        int lo=k-1, hi=n;
        while(lo<hi){
            int m=(lo+hi)/2; 
            // cerr << m << endl;
            if(qry(m)+sum>2*a) hi=m-1; 
            else if(qry(m)+sum<a) lo=m+1; 
            else{
                vector<int> Ans; F0R(i, k-1) Ans.pb(i+1);
                Ans.pb(m+1); 
                answer(Ans); exit(0);
            }
        }
    }
    int lo=0, hi=n; 
    while(lo+1<hi){
        int m=(lo+hi)/2; 
        if(qry(m)*k>=a) hi=m;
        else lo=m+1; 
    }
    FOR(i, lo-2*k+1, lo+k+1){
        chk(i);
    }
    impossible();
    // assert(n==s);
    // task is query 40 books and try to see if we can get a set of K books such that A<=\sum<=2A we have K is less than 10 so SMALL 
    // solve subtask by subtask:
    /*
        observation, problem has to require BINARY SEARCH
        subtask 1: we can just like iterte over first 2 books, check which if we have a third one 
        subtask 2: if the last element we take is <=A then if the previous elements are >A, we can find a solution 

        current solution:
        binary search for first X such that x[X]*K>=A then notice this element has to be used
        at least I think that is true 

        20+17 queries minus a couple 

        thing is we also need to check for the smallest number with value >= A 

        We probably have 80 points 
    */
    // vector<ll> x(n); 
    // F0R(i, n){
    //     x[i]=skim(i+1); 
    // }
    // // {
    // //     ll p=0; 
    // //     F0R(i, k-1) p+=x[i]; 
    // //     FOR(i, k-1, n){
    // //         if(p+x[i]>=A&&2*A>=p+x[i]){
    // //             vector<int> ans; 
    // //             F0R(i, k-1) ans.pb(i+1); 
    // //             ans.pb(i+1); 
    // //             answer(ans); 
    // //         }
    // //     }
    // // }
    // ll sum=0; 
    // F0R(i, k-1) sum+=x[i]; 
    // FOR(i, k-1, n){
    //     sum+=x[i]; 
    //     if(sum>=a&&sum<=2*a){
    //         vector<int> ans;
    //         FOR(j, i-k+1, i+1)  ans.pb(j+1); 
    //         answer(ans); 
    //     }
    //     sum-=x[i-k+1];
    // }
    // impossible();
}

// int main(){
//     cin.tie(0)->sync_with_stdio(0); 

// }

• 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