이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "boxes.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff(i,a,b) for(int i=a;i<=b;i++)
#define fb(i,b,a) for(int i=b;i>=a;i--)
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 100005;
const ll inf = 1e18 + 5;
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k) print the k-th smallest number in os(0-based)
int n, k, l;
int niz[maxn];
ll dp1[maxn];
ll dp2[maxn];
ll delivery(int N, int K, int L, int positions[]){
n = N, k = K, l = L;
ff(i,1,n)niz[i] = positions[i - 1];
ff(i,1,n)dp1[i] = dp2[i] = inf;
dp1[0] = 0;
ff(i,1,n)dp1[i] = dp1[max(0, i - k)] + 2 * niz[i];
dp2[n + 1] = 0;
fb(i,n,1)dp2[i] = dp2[min(i + k, n + 1)] + 2 * (l - niz[i]);
ll rez = inf;
ff(i,0,n)rez = min(rez, dp1[i] + dp2[i + 1]);
ff(i,1,n)rez = min(rez, dp1[i] + dp2[min(i + k + 1, n + 1)] + l);
return rez;
}
//int main()
//{
// ios::sync_with_stdio(false);
// cout.tie(nullptr);
// cin.tie(nullptr);
// cin >> n >> k >> l;
// ff(i,1,n)cin >> niz[i];
//
// ff(i,1,n)dp1[i] = dp2[i] = inf;
//
// dp1[0] = 0;
// ff(i,1,n)dp1[i] = dp1[max(0, i - k)] + 2 * niz[i];
//
// dp2[n + 1] = 0;
// fb(i,n,1)dp2[i] = dp2[min(i + k, n + 1)] + 2 * (l - niz[i]);
//
// ll rez = inf;
// ff(i,0,n)rez = min(rez, dp1[i] + dp2[i + 1]);
// ff(i,1,n)rez = min(rez, dp1[i] + dp2[min(i + k + 1, n + 1)] + l);
// cout << rez << endl;
//
// return 0;
//}
/**
3 2 8
1 2 5
// probati bojenje sahovski ili slicno
**/
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