# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
1049317 | SN0WM4N | 순열 (APIO22_perm) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define sort undefined_function // To use stable_sort instead sort
#define bpc __builtin_popcount
#define ull unsigned long long
#define ld double
#define ll long long
#define mp make_pair
#define F first
#define S second
#pragma GCC optimize("O3")
#ifdef LOCAL
#include "debug.h"
#else
#define dbg(...) 0
#include "perm.h"
#endif
using namespace __gnu_pbds;
using namespace std;
typedef tree<long long, null_type, less_equal<long long>,
rb_tree_tag, tree_order_statistics_node_update> Tree;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll INF = 9223372036854775807LL;
const ll inf = 2147483647;
const ll MOD = 998244353; //[998244353, 1e9 + 7, 1e9 + 13]
const ld PI = acos(-1);
const ll NROOT = 800;
ll binpow(ll a, ll b, ll _MOD = -1) {
if (_MOD == -1)
_MOD = MOD;
ll res = 1;
for (; b; b /= 2, a *= a, a %= _MOD)
if (b & 1) res *= a, res %= _MOD;
return res;
}
void set_IO(string s) {
#ifndef LOCAL
string in = s + ".in";
string out = s + ".out";
freopen(in.c_str(), "r", stdin);
freopen(out.c_str(), "w", stdout);
#endif
}
bool dataOverflow(ll a, ll b) {return (log10(a) + log10(b) >= 18);}
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a * b / gcd(a, b);}
ll ceil(ll a, ll b) {return (a + b - 1) / b;}
ll invmod(ll a) {return binpow(a, MOD - 2);}
vector<int> construct_permutation(ll k) {
ll size = log2(k);
vector<ll> ans;
vector<ll> aux;
for (ll i = 100; i >= 0; i --)
aux.push_back(100000000 - i);
for (ll i = 0; i < size; i ++) {
if (k & (1 << i)) {
ans.push_back(aux.back());
aux.pop_back();
}
ans.push_back(i);
}
set<ll> s;
for (auto &x : ans)
s.insert(x);
int cnt = 0;
map<ll, ll> ID;
for (auto &x : s)
ID[x] = cnt ++;
for (auto &x : ans)
x = ID[x];
return ans;
}