結果

問題 No.2844 Birthday Party Decoration
ユーザー 👑 binapbinap
提出日時 2024-08-23 21:47:07
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 869 ms / 2,000 ms
コード長 3,318 bytes
コンパイル時間 4,612 ms
コンパイル使用メモリ 262,968 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-08-23 21:47:15
合計ジャッジ時間 8,624 ms
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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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
6,816 KB
testcase_01 AC 804 ms
6,940 KB
testcase_02 AC 827 ms
6,944 KB
testcase_03 AC 868 ms
6,940 KB
testcase_04 AC 869 ms
6,940 KB
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 #
プレゼンテーションモードにする

#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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