Submission #601508

#	Time	Username	Problem	Language	Result	Execution time	Memory
601508		thezomb1e	Permutation (APIO22_perm)	C++17	64.63 / 100	14 ms	1492 KiB

perm

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

#define eb emplace_back
#define pb push_back
#define ft first
#define sd second
#define pi pair<int, int>
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define dbg(...) dbg_out(__VA_ARGS__)

using ll = long long;
using ld = long double;
using namespace std;
using namespace __gnu_pbds;

//Constants
const ll INF = 5 * 1e18;
const int IINF = 2 * 1e9;
const ll MOD = 1e9 + 7;
// const ll MOD = 998244353;
const ll dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ld PI = 3.14159265359;

//Templates
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) {return os << '(' << p.first << ", " << p.second << ')';}
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) {os << '['; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << ']';}
void dbg_out() {cerr << endl;}
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << H << ' '; dbg_out(T...); }
template<typename T> void mins(T& x, T y) {x = min(x, y);}
template<typename T> void maxs(T& x, T y) {x = max(x, y);}
template<typename T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using omset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;


vector<int> construct_permutation(ll k) {
	if (!(k & (k - 1))) {
		for (int i = 0; i < 60; i++) {
			if (k & (1ll << i)) {
				vector<int> ans;
				for (int j = 0; j < i; j++) {
					ans.pb(j);
				}
				return ans;
			}
		}
	}
    k--;
	int pp = 0, add = 0;
	vector<int> ans;
	for (int i = 0; i < 60; i++) {
		if (k & (1ll << i)) {
			for (int j = pp + i - 1; j >= pp; j--) {
				ans.pb(j);
			}
            add++;
			pp += i;
		}
	}
    for (int i = 0; i < add; i++) {
        ans.pb(pp++);
    }
	reverse(all(ans));
	return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 54.63076923 / 90