Submission #1167442

#	Time	Username	Problem	Language	Result	Execution time	Memory
1167442		pan	Liteh and Newfiteh (INOI20_litehfiteh)	C++20	100 / 100	72 ms	41136 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll const INF = 1e18;
ll a[100005], dp[100005][25], dp2[100005][25], dp3[100005];

void solve()
{
	ll n, m;
	cin >> n;
	m = 0;
	while (1LL << (m + 1) <= n) m++;
	for (ll i=1; i<=n; ++i) 
	{
		cin >> a[i];
		if (a[i] > m) {cout << "-1\n"; return;}
	}
	for (ll i=1; i<=n; ++i) for (ll j=0; j<=m + 1; ++j) dp[i][j] = INF, dp2[i][j] = INF;
	for (ll j=1; j<=n;++j) dp[j][a[j]] = 0;
	for (ll i=0; i<= m; ++i)
	{
		// merge two previous layer
		if (i)
		{
			for (ll j=1; j<=n - (1LL << i) + 1; ++j) 
				for (ll v=0; v<=m; ++v) 
					dp[j][v] = min(INF, dp[j][v] + dp[j + (1LL << (i-1))][v]);
		}
		// add or not add
		for (ll j=1; j<=n - (1LL << i) + 1; ++j) for (ll v=0; v<=m; ++v) dp[j][v] = min(dp[j][v], dp[j][v+1] + 1);
		// add results 
		for (ll j=1; j<=n - (1LL << i) + 1; ++j) dp2[j][i] = dp[j][0];
	}
	dp3[0] = 0;
	for (ll i=1; i<=n; ++i) dp3[i] = INF;
	for (ll i=1; i<=n; ++i) for (ll j=0; j<=m && i +  (1LL << j) -1 <= n; ++j) dp3[i + (1LL << j)-1] = min(dp3[i + (1LL << j)-1], dp3[i-1] + dp2[i][j]);
	cout << ((dp3[n]>=INF)?-1:dp3[n]) << endl;
}

int main()
{
	cin.tie(nullptr);
	cout.tie(nullptr);
	ios::sync_with_stdio(false);
	ll t = 1;
	while (t--) solve();
	return 0;
}

• Subtask 1: 30 / 30
• Subtask 2: 70 / 70