Submission #751516

#	Time	Username	Problem	Language	Result	Execution time	Memory
751516		GrindMachine	Mountains (IOI17_mountains)	C++17	100 / 100	25 ms	16084 KiB

mountains

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://oj.uz/submission/684363

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
const ld eps = 1e-9;

#include "mountains.h"

int maximum_deevs(vector<int> a) {
	int n = sz(a);
	int dp[n+5][n+5];
	memset(dp,0,sizeof dp);

	rev(l,n-1,0){		
		dp[l][l] = 1;

		ld grad = -inf2;
		int recent = l;
		int sum = 1;

		for(int r = l + 1; r < n; ++r){
			// ignore l
			dp[l][r] = dp[l+1][r];

			// pick l
			ld curr = (a[r]-a[l]) / (ld) (r-l);
			if(curr >= grad-eps){
				sum += dp[recent+1][r-1];
				recent = r;
				grad = curr;
			}

			sum += dp[recent+1][r];
			amax(dp[l][r], sum);

			sum -= dp[recent+1][r];
		}
	}

	// rep(l,n){
	// 	rep(r,n){
	// 		cout << dp[l][r] << " ";
	// 	}
	// 	cout << endl;
	// }

	return dp[0][n-1];
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 30 / 30
• Subtask 4: 30 / 30