Submission #90478

#	Time	Username	Problem	Language	Result	Execution time	Memory
90478		radoslav11	Brunhilda’s Birthday (BOI13_brunhilda)	C++14	17.46 / 100	1095 ms	158760 KiB

brunhilda

/*
	There is a simple solution with upper bound on the time equal to O(MAX * log(log(MAX)) * log(MAX)) with DP and maintaining a heap. Probably the actual complexity is smaller.
*/

#include <bits/stdc++.h>
#define endl '\n'

//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")

#define SZ(x) ((int)x.size())
#define ALL(V) V.begin(), V.end()
#define L_B lower_bound
#define U_B upper_bound
#define pb push_back

using namespace std;
template<class T, class T2> inline int chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline int chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const int MAXN = (1 << 20);
const int B = (int)1e7 + 42;
const int inf = B + 42;

int n, q;
int dp[B + 42], cnt[inf + 42];
int a[MAXN], lp[B + 42];

void read()
{
	cin >> n >> q;
	for(int i = 0; i < n; i++)
		cin >> a[i];
}

priority_queue<int, vector<int>, greater<int> > Q;
int last[MAXN];

void fix()
{
	while(!Q.empty() && cnt[Q.top()])
	{
		cnt[Q.top()]--;
		Q.pop();
	}
}

void solve()
{
	for(int i = 0; i < a[n - 1]; i++) dp[i] = 1;
	for(int i = a[n - 1]; i <= B; i++) dp[i] = inf;

	memset(lp, -1, sizeof(lp));
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j <= B; j += a[i])
			if(dp[j] != 1 && lp[j] == -1)
				lp[j] = i;
	}

	for(int i = 0; i < n - 1; i++)
		Q.push(1), last[i] = 1;

	forward_list<int> li;
	for(int d = a[n - 1]; d <= B; d++)
	{
		li.clear();

		int dummy = d;
		while(lp[dummy] != -1)
		{
			int D = a[lp[dummy]];
			li.push_front(lp[dummy]);
			while(dummy % D == 0) dummy /= D;
		}

		for(int i: li) cnt[last[i]]++;
		
		fix();	
		if(!Q.empty()) 
			chkmin(dp[d], Q.top() + 1);
	
		for(int i: li)
		{
			Q.push(dp[d]);
			last[i] = dp[d];
		}
	}

	while(q--)
	{
		int x;
		cin >> x;
		if(dp[x] == inf) cout << "oo" << endl;
		else cout << dp[x] << endl;
	}
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	read();
	solve();
	return 0;
}

• Subtask 1: 4.444444444444445 / 20
• Subtask 2: 4.444444444444445 / 20
• Subtask 3: 8.571428571428571 / 60