結果

問題 No.3081 Make Palindromic Multiple
コンテスト
ユーザー 👑 binap
提出日時 2025-02-24 02:34:47
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 13 ms / 1,000 ms
コード長 4,319 bytes
コンパイル時間 4,538 ms
コンパイル使用メモリ 257,468 KB
実行使用メモリ 7,904 KB
最終ジャッジ日時 2025-03-27 13:22:58
合計ジャッジ時間 8,129 ms
ジャッジサーバーID
（参考情報） 		judge3 / 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 * 3
other AC * 54
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;

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 (long long i = 2; i <= n; ++i) {
      if (f[i]) continue;
      primes.push_back(i);
      f[i] = i;
      for (long long 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<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 n;
	cin >> n;
	
	Sieve sieve(1000005);
	auto factor = sieve.factor(n);
	
	int two = 0;
	int five = 0;
	
	for(auto [key, val] : factor){
		if(key == 2) two = val;
		if(key == 5) five = val;
	}
	
	assert(two == 0 or five == 0);
	
	// n = 2^e2 * 5^e5 * m
	long long m = n;
	
	string s;
	if(two > 0){
		long long x = 1;
		rep(i, two) x *= 2;
		
		m /= x;
		
		string s_tmp = to_string(x);
		int sz = s_tmp.size();
		
		while(sz < two){
			s_tmp.insert(0, "0");
			sz++;
		}
		
		for(int i = sz - 1; i >= 0; i--) s += s_tmp[i];
		for(int i = 0; i < sz; i++) s += s_tmp[i];
	}else if(five > 0){
		long long x = 1;
		rep(i, five) x *= 5;
		
		m /= x;
		
		string s_tmp = to_string(x);
		int sz = s_tmp.size();
		
		while(sz < five){
			s_tmp.insert(0, "0");
			sz++;
		}
		
		for(int i = sz - 1; i >= 0; i--) s += s_tmp[i];
		for(int i = 0; i < sz; i++) s += s_tmp[i];
		
	}else{
		s = "1";
	}
	
	{
		for(auto [p, e] : factor){
			if(p == 2) continue;
			if(p == 5) continue;
			
			// 10^|S| mod N
			long long ten = 1;
			rep(i, int(s.size())){
				ten *= 10;
				ten %= p;
			}
			
			if(ten != 1){
				m /= p;
				m *= p - 1;
			}
		}
	}
	
	cout << 1 << "\n";
	cout << s << ' ' << m << "\n";
	
	return 0;
}
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