Submission #415271

#	Time	Username	Problem	Language	Result	Execution time	Memory
415271		Carmel_Ab1	Detecting Molecules (IOI16_molecules)	C++17	100 / 100	91 ms	10180 KiB

molecules

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

//#include "grader.cpp"

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef vector<ll>vl;
typedef vector<vl> vvl;
typedef pair<int,int>pi;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld,ld> pld;
//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;}

#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map

template<typename T1, typename T2> istream& operator>>(istream& is, pair<T1, T2>& p){is >> p.first >> p.second;return is;}
template<typename T1, typename T2> ostream& operator<<(ostream& os, pair<T1, T2>& p){os <<"" << p.first << " " << p.second << ""; return os;}

vi find_subset(int L, int U, vi W){
    ll l=ll(L),u=ll(U),n=W.size();
    vl w(n);
    for(int i=0; i<n; i++)
        w[i]=ll(W[i]);
    vpl a(n);
    for(int i=0; i<n; i++)
        a[i]={w[i],i};
    sort(all(a));
    vl pref(n);
    for(int i=0; i<n; i++)
        pref[i]=(i?pref[i-1]:0)+a[i].first;

    for(int i=0; i<n; i++){
        int li=i,ri=n-1,j=-1;

        while(li<=ri){
            int mi=(li+ri)/2;
            if(pref[mi]-(i?pref[i-1]:0)<=u)
                j=mi,li=mi+1;
            else
                ri=mi-1;
        }
        if(j==-1 || pref[j]-(i?pref[i-1]:0)<l)continue;
        vi ans;
        for(int k=i;k<=j; k++)
            ans.pb(a[k].second);
        sort(all(ans));
        return ans;
    }
    return {};
}

• Subtask 1: 9 / 9
• Subtask 2: 10 / 10
• Subtask 3: 12 / 12
• Subtask 4: 15 / 15
• Subtask 5: 23 / 23
• Subtask 6: 31 / 31