結果

問題 No.3081 Make Palindromic Multiple
ユーザー 👑 binap
提出日時 2025-03-01 02:32:50
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 5,482 bytes
コンパイル時間 5,056 ms
コンパイル使用メモリ 264,260 KB
実行使用メモリ 9,804 KB
最終ジャッジ日時 2025-03-27 13:25:34
合計ジャッジ時間 27,122 ms
ジャッジサーバーID
（参考情報） 		judge1 / 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 * 15 WA * 39
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;
  }
};

template<typename T>
T pow_mod(T A, T N, T MOD){
	assert(N >= 0);
	if(N == 0) return T(1) % MOD;
	if(N == 1) return T(A) % MOD;
	A %= MOD;
	if(N % 2 == 0){
		auto p = pow_mod<T>(A, N / 2, MOD);
		T p2 = (p * p) % MOD;
		return p2;
	}else{
		auto p = pow_mod<T>(A, N - 1, MOD);
		T p2 = (p * A) % MOD;
		return p2;
	}
}

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;
		
		s = to_string(x);
		
	}else if(five > 0){
		long long x = 1;
		rep(i, five) x *= 5;
		
		m /= x;
		
		s = to_string(x);
	}else{
		s = "1";
	}
	
	long long phi = m;
	for(auto [p, e] : factor){
		if(p == 2) continue;
		if(p == 5) continue;
		phi /= p;
		phi *= p - 1;
	}
	
	auto solve = [&](long long d){
		long long remain = 0;
		{
			remain += stoll(s);
		}
		{
			reverse(s.begin(), s.end());
			long long x = stoll(s);
			remain += x * pow_mod<__uint128_t>(10, 2 * d + 1 - s.size(), m);
			reverse(s.begin(), s.end());
		}
		{
			remain = (m + (-remain) % m) % m;
		}
		vector<pair<long long, int>> vec;
		{
			vec.emplace_back(1LL, 0);
		}
		int dig = min(d / 2, 80000LL);
		for(int idx = 1; idx <= dig; idx++){
			long long c1 = pow_mod<__uint128_t>(10, d + idx, m);
			long long c2 = pow_mod<__uint128_t>(10, d - idx, m);
			long long c = (c1 + c2) % m;
			vec.emplace_back(c, idx);
		}
		sort(vec.rbegin(), vec.rend());
		
		string t;
		rep(i, 2 * dig + 1) t += '0';
		for(auto [val, idx] : vec){
			for(int num = 9; num >= 1; num--){
				if(val == 0) continue;
				if(remain >= val * num){
//					cout << remain << ' ' << val << ' ' << num << ' ' << idx << "\n";
					remain -= val * num;
					t[dig + idx] = '0' + num;
					t[dig - idx] = '0' + num;
					break;
				}
			}
		}
		return make_pair(dig, t);
	};
	auto[dig, t] = solve(phi * 100);
	int sz = s.size();
	cout << 5 << "\n";
	{
		reverse(s.begin(), s.end());
		cout << s << ' ' << "1\n";
		reverse(s.begin(), s.end());
	}
	{
		cout << '0' << ' ' << phi * 100 - dig - sz << "\n";
	}
	{
		cout << t << ' ' << "1\n";
	}
	{
		cout << '0' << ' ' << phi * 100 - dig - sz << "\n";
	}
	{
		cout << s << ' ' << "1\n";
	}
	return 0;
}
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