結果

問題 No.2844 Birthday Party Decoration
コンテスト
ユーザー 👑 binap
提出日時 2024-08-23 21:47:07
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 849 ms / 2,000 ms
コード長 3,318 bytes
コンパイル時間 4,207 ms
コンパイル使用メモリ 251,864 KB
最終ジャッジ日時 2025-02-23 23:39:28
ジャッジサーバーID
（参考情報） 		judge2 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 4
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;

ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

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

int solve(){
	int n;
	long long x;
	cin >> n >> x;
	vector<int> c(n);
	cin >> c;
	vector<long long> l_vec(n + 1);
	rep(i, n){
		long long p = x / (1LL << (c[i] + 1));
		long long q = x % (1LL << (c[i] + 1));
		if(q >= (1LL << c[i])) l_vec[i] = x;
		else{
			l_vec[i] = p * (1LL << (c[i] + 1)) - 1;
		}
	}
	l_vec[n] = x;
	vector<long long> r_vec(n + 1);
	rep(i, n){
		long long p = x / (1LL << (c[i] + 1));
		long long q = x % (1LL << (c[i] + 1));
		if(q >= (1LL << c[i])) r_vec[i] = x;
		else{
			r_vec[i] = p * (1LL << (c[i] + 1)) + (1LL << c[i]);
		}
	}
	r_vec[n] = x;
//	cout << l_vec;
//	cout << r_vec;
	long long ans = (1LL << 62);
	for(long long l : l_vec){
		if(l == -1) continue;
		for(long long r : r_vec){
			assert(l <= r);
			long long d = r - l;
			bool success = true;
			rep(i, n){
				if(d >= (1LL << c[i])) continue;
				long long q_l = l % (1LL << (c[i] + 1));
				long long q_r = r % (1LL << (c[i] + 1));
				if(q_l < (1LL << c[i]) and q_r < (1LL << c[i])) success = false;
			}
			if(success) chmin(ans, (r - l) * 2);
//			cout << l << ' ' << r << ' ' << ans << "\n";
		}
	}
	cout << ans << "\n";
	return 0;
}

int main(){
	int t;
	cin >> t;
	rep(_, t) solve();
	return 0;
}
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