結果

問題 No.3028 No.9999
ユーザー 👑 binap
提出日時 2025-02-21 21:36:40
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 24 ms / 4,000 ms
コード長 4,058 bytes
コンパイル時間 4,374 ms
コンパイル使用メモリ 259,840 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2025-02-21 21:36:48
合計ジャッジ時間 5,360 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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 * 23
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;
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 num_of_divisors(long long x){
        auto f = factor(x);
        long long res = 1;
        for(auto [p, val] : f){
            res *= val + 1;
        }
        return res;
    }
    set<long long> divisors(long long x){
        set<long long> se = {1};
        auto f = factor(x);
        for(auto [key, value] : f){
            set<long long> old = se;
            for(long long y : old){
                rep(i, value + 1){
                    se.insert(y);
                    y *= key;
                }
            }
        }
        return se;
    }
};
template<typename T>
T pow_mod(T A, T N, T MOD){
    T res = 1 % MOD;
    A %= MOD;
    while(N){
        if(N & 1) res = (res * A) % MOD;
        A = (A * A) % MOD;
        N >>= 1;
    }
    return res;
}
int main(){
    int n;
    cin >> n;
    
    if(n == 1){
        cout << "1\n";
        return 0;
    }
    
    Sieve sieve(10005);
    auto factor = sieve.factor(n);
    
    int phi = n;
    for(auto [p, cnt] : factor){
        phi /= p;
        phi *= p - 1;
    }
    
    auto list = sieve.divisors(phi);
    for(auto e : list){
        auto res = pow_mod<long long>(10, e, n);
        if(res == 1){
            cout << e << "\n";
            return 0;
        }
    }
    
    
    
    return 0;
}
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