結果

問題 No.3127 Multiple of Twin Prime
ユーザー 👑 binap
提出日時 2025-04-25 21:42:19
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 553 ms / 2,500 ms
コード長 2,636 bytes
コンパイル時間 4,623 ms
コンパイル使用メモリ 252,772 KB
実行使用メモリ 42,888 KB
最終ジャッジ日時 2025-04-25 21:42:33
合計ジャッジ時間 11,736 ms
ジャッジサーバーID
(参考情報)
judge2 / judge4
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 * 12
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 pair<int, int> P;

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 <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
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);}

const int M = 10000000;

int main(){
	vector<int> f(M + 1);
	vector<long long> g;
	
	for(int x = 2; x <= M; x++){
		if(f[x] == 1) continue;
		for(int y = 2 * x; y <= M; y += x) f[y] = 1;
	}
	
	for(int x = 2; x + 2 <= M; x++){
		if(f[x] == 0 and f[x + 2] == 0) g.push_back((long long)x * (x + 2));
	}
	
	int q;
	cin >> q;
	rep(_, q){
		long long x;
		cin >> x;
		auto itr = upper_bound(g.begin(), g.end(), x);
		if(itr == g.begin()){
			cout << "-1\n";
		}else{
			cout << *prev(itr) << "\n";
		}
	}
	
	return 0;
}
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