Submission #336825

#	Time	Username	Problem	Language	Result	Execution time	Memory
336825		shrek12357	Detecting Molecules (IOI16_molecules)	C++14	100 / 100	56 ms	5612 KiB

molecules

#include "molecules.h"
#include<bits/stdc++.h>
 
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<ll, ll>> vpll;
 
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)
 
#define pb push_back
#define mp make_pair
#define rsz resize
#define sz(x) int(x.size())
#define all(x) x.begin(),x.end()
#define f first
#define s second
#define endl '\n'
#define test int testc;cin>>testc;while(testc--)
const int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 };
const ll linf = 4000000000000000000LL;
const ll inf = 1000000007;
 
void pv(vi a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vll a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vector<vi>a) {
	F0R(i, sz(a)) { cout << i << endl; pv(a[i]); cout << endl; }
}void pv(vector<vll>a) { F0R(i, sz(a)) { cout << i << endl; pv(a[i]); }cout << endl; }void pv(vector<string>a) { trav(x, a)cout << x << endl; cout << endl; }
 
 
vi find_subset(int l, int r, vi w)
{
	int n = w.size();
	vpi a(n);
	F0R(i, n)a[i] = { w[i],i };
	sort(all(a));
	ll prefix = 0, suffix = 0;
	F0R(i, n)
	{
		prefix += a[i].f;
		suffix += a[n - i - 1].f;
		if (prefix >= l && prefix <= r)
		{
			vi ans;
			F0R(j, i + 1)
			{
				ans.pb(a[j].s);
			}
			return ans;
		}
		else if (suffix >= l && suffix <= r)
		{
			vi ans;
			F0R(j, i + 1)
			{
				ans.pb(a[n - j - 1].s);
			}
			return ans;
		}
		else
		{
			if (prefix > r && suffix > r)
			{
				vi ans(0);
				return ans;
			}
			if (prefix < l && suffix > r)
			{
				ll sum = 0;
				F0R(j, i + 1)sum += a[j].f;
				int left = 0, right = i;
				while (right < n - 1)
				{
					sum -= a[left++].f;
					sum += a[++right].f;
					if (sum >= l && sum <= r)
					{
						vi ans;
						FOR(j, left, right + 1)ans.pb(a[j].s);
						return ans;
					}
				}
				assert(0);
			}
		}
	}
	vi ans(0);
	return ans;
}

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