제출 #415356

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
415356		kym	Detecting Molecules (IOI16_molecules)	C++14	0 / 100	1 ms	204 KiB

molecules

#include <bits/stdc++.h>
using namespace std;
//#define int ll 
#define FOR(i,s,e) for(ll i = s; i <= (ll)e; ++i)
#define DEC(i,s,e) for(ll i = s; i >= (ll)e; --i)
#define IAMSPEED ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef LOCAL
#define db(x) cerr << #x << "=" << x << "\n"
#define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << "\n"
#define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<"\n"
#define dbv(v) cerr << #v << ":"; for (auto ite : v) cerr << ite << ' '; cerr <<"\n"
#define dbvp(v) cerr << #v << ":"; for (auto ite : v) cerr << "{"  << ite.f << ',' << ite.s << "} "; cerr << "\n"
#define dba(a,ss,ee) cerr << #a << ":"; FOR(ite,ss,ee) cerr << a[ite] << ' '; cerr << "\n"
#define reach cerr << "LINE: " << __LINE__ << "\n";
#else
#define db(x)
#define db2(x,y)
#define db3(a,b,c)
#define dbv(v)
#define dbvp(v)
#define dba(a,ss,ee)
#define reach 
#endif
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define ll long long 
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define f first
#define s second
#define g0(x) get<0>(x)
#define g1(x) get<1>(x)
#define g2(x) get<2>(x)
#define g3(x) get<3>(x)
typedef pair <ll, ll> pi;
typedef tuple<ll,ll,ll> ti3;
typedef tuple<ll,ll,ll,ll> ti4;
ll rand(ll a, ll b) { return a + rng() % (b-a+1); }
const int MOD = 1e9 + 7;
const int inf = (int)1e9 + 500;
const long long oo = (ll)1e18 + 500;
template <typename T> bool chmax(T& a, const T b) { return a<b ? a = b, 1 : 0; }
template <typename T> bool chmin(T& a, const T b) { return a>b ? a = b, 1 : 0; }
const int MAXN = 200005;

//#include "molecules.h"
ll A[MAXN], sum[MAXN];
int n, lbd, ubd;

vector<int> done;
bool gotans=0;
bool check(int numtake) {
	db(numtake);
	int ss = 0;
	FOR(i,1,numtake) ss+=A[i]-A[1];
	ss+=A[1]*numtake;
	db(ss);
	if(ss>ubd)return 0;
	if(ss>=lbd){
		FOR(i,1,numtake) {
			done.pb(i);
		}
		gotans=1;
		return 1;
	}
	FOR(i,1,numtake) {
		ss-=A[i]-A[1];
		ss+=A[n-numtake+i]-A[1];
		if(ss>=lbd){
			FOR(j,i+1,numtake)done.pb(j);
			FOR(j,n-numtake+1,n-numtake+i)done.pb(j);
			gotans=1;
			return 1;
		}
		assert(ss<=ubd);
	}
	return 1;
}
vector<int> ww;

bool comp(int x, int y) {
	return ww[x]<ww[y];
}

vector<int> find_subset(int l, int u, vector<int> w) {
	gotans=0;
	ww = w;
	vector<ll> ord;
	n = w.size();
	FOR(i,0,n-1) {
		ord.pb(i);
	}
	sort(all(ord), comp);
	lbd=l; ubd=u;
	sort(all(w));
	FOR(i,0,n-1) {
		A[i+1]=w[i];
	}
	int lo = 0, hi = n+1;
	while(lo<hi-1) {
		int mid=(lo+hi)/2;
		if(check(mid)) {
			if(gotans) goto imdone;
			lo=mid;
		}
		hi = mid;
	}
	imdone:;
	vector<int> ans;
	FOR(i,0,done.size()-1) {
		ans.pb(ord[done[i]-1]);
	}
    return ans;
}

• Subtask 1: 0 / 9
• Subtask 2: 0 / 10
• Subtask 3: 0 / 12
• Subtask 4: 0 / 15
• Subtask 5: 0 / 23
• Subtask 6: 0 / 31