Submission #692313

#	Time	Username	Problem	Language	Result	Execution time	Memory
692313		saayan007	Mountains (IOI17_mountains)	C++17	0 / 100	1 ms	212 KiB

mountains

#include "mountains.h"
#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using pi = pair<int, int>;
using pl = pair<long long, long long>;
using vi = vector<int>;
using vl = vector<long long>;
using vpi = vector<pair<int, int>>;
using vpl = vector<pair<long long, long long>>;

#define fur(i, a, b) for(ll i = a; i <= (ll)b; ++i)
#define ruf(i, a, b) for(ll i = a; i >= (ll)b; --i)
#define fr first 
#define sc second
#define mp make_pair
#define pb emplace_back
#define nl "\n"
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()

bool check(vector<int> &y, int i, int j, int k) {
    int A = y[i] - y[j];
    int B = i - j;
    int C = y[i] * (j - i) + i * (y[i] - y[j]);

    return A*k + B*y[k] + C > 0;
}

int maximum_deevs(vector<int> y) {
    int ans = 1;
    int n = y.size();

    for(int b = 0; b < (1 << n); ++b) {
        bool valid = 1;
        int count = 0;
        for(int i = 0; i < n; ++i) {
            if(!(b & (1 << i)))
                continue;
            ++count;
            for(int j = i + 1; j < n; ++j) {
                if(!(b & (1 << j)))
                    continue;
                bool block = 0;
                for(int k = i + 1; k < j; ++k) {
                    block |= check(y, i, j, k);
                }
                valid &= block;
            }
        }
        if(valid && count > ans) {
            ans = count;
        }
    }
    return ans;
    // bool ok[n][n] = {};
    // for(int i = 0; i < n; ++i) {
        // for(int j = i + 2; j < n; ++j) {
            // for(int k = i + 1; k < j; ++k) {
                // ok[i][j] = check(y, i, j, k);
            // }
        // }
    // }

    // int dp[n] = {};
    // dp[n] = 1;
    // for(int i = n - 1; i >= 0; --i) {
        // dp[i] = 1;
        // for(int j = i + 2; j < n; ++j) {
            // if(ok[i][j]) {
                // dp[i] = max(dp[i], dp[j] + 1);
            // }
        // }
        // ans = max(ans, dp[i]);
    // }

    // return ans;
}

• Subtask 1: 0 / 20
• Subtask 2: 0 / 20
• Subtask 3: 0 / 30
• Subtask 4: 0 / 30