#1148317 (C++17) No.3485 Find 495-like Number

• 問題: No.3485 Find 495-like Number
• ユーザー: 👑 binap
• 提出日時: 2026-02-16 06:51:47
• 言語: C++17 (gcc 15.2.0 + boost 1.89.0)
• 結果: AC
• 実行時間: 1,557 ms / 5,000 ms
• コード長: 4,036 bytes
• コンパイル時間: 4,979 ms
• コンパイル使用メモリ: 279,868 KB
• 実行使用メモリ: 101,480 KB
• 最終ジャッジ日時: 2026-03-27 20:55:40
• 合計ジャッジ時間: 35,535 ms
• ジャッジサーバーID （参考情報）: judge3_1 / judge2_0

ファイルパターン	結果
sample	AC * 3
other	AC * 34

ソースコード

// https://t5k.org/notes/GapsTable.html

#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);}

// Sieve of Eratosthenes
// https://youtu.be/UTVg7wzMWQc?t=2774
struct Sieve {
  int n;
  vector<int> f, primes;
  Sieve(int n=1):n(n), f(n+1) {
    f[0] = f[1] = -1;
    for (ll i = 2; i <= n; ++i) {
      if (f[i]) continue;
      primes.push_back(i);
      f[i] = i;
      for (ll j = i*i; j <= n; j += i) {
        if (!f[j]) f[j] = i;
      }
    }
  }
  bool isPrime(int x) { return f[x] == x;}
  vector<int> factorList(int x) {
    vector<int> res;
    while (x != 1) {
      res.push_back(f[x]);
      x /= f[x];
    }
    return res;
  }
  vector<P> factor(int x) {
    vector<int> fl = factorList(x);
    if (fl.size() == 0) return {};
    vector<P> res(1, P(fl[0], 0));
    for (int p : fl) {
      if (res.back().first == p) {
        res.back().second++;
      } else {
        res.emplace_back(p, 1);
      }
    }
    return res;
  }
  vector<pair<ll,int>> factor(ll x) {
    vector<pair<ll,int>> res;
    for (int p : primes) {
      int y = 0;
      while (x%p == 0) x /= p, ++y;
      if (y != 0) res.emplace_back(p,y);
    }
    if (x != 1) res.emplace_back(x,1);
    return res;
  }
};

int main(){
	long long l, r;
	cin >> l >> r;
	Sieve sieve(22250000);
	auto primes = sieve.primes;
	auto judge = [&](long long x){
		vector<pair<long long, int>> factors;
		for(auto p : primes){
			int cnt = 0;
			while(x % p == 0){
				cnt++;
				x/= p;
			}
			if(cnt > 0) factors.emplace_back(p, cnt);
		}
		if(x > 1) factors.emplace_back(x, 1);
		if(factors.size() == 3){
			if(factors[0].first != 2 and factors[0].second == 2 and factors[1].second == 1 and factors[2].second == 1){
				return 1;
			}
		}
		return 0;
	};
	for(long long x = ((l + 44) / 45 * 45); x <= r; x += 45){
		if(judge(x)){
			cout << x << "\n";
			return 0;
		}
	}
	for(long long x = l; x <= r; x++){
		if(judge(x)){
			cout << x << "\n";
			return 0;
		}
	}
	cout << "-1\n";
	return 0;
}