#886044 (C++17) No.2365 Present of good number

• 問題: No.2365 Present of good number
• ユーザー: pointN
• 提出日時: 2023-06-30 22:18:01
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 197 ms / 2,000 ms
• コード長: 2,557 bytes
• コンパイル時間: 1,795 ms
• コンパイル使用メモリ: 145,424 KB
• 最終ジャッジ日時: 2025-02-15 04:06:11
• ジャッジサーバーID （参考情報）: judge4 / judge3

ファイルパターン	結果
sample	AC * 2
other	AC * 39

ソースコード

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#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <iomanip>
#include <stack>
#include <queue>
#include <numeric>
#include <map>
#include <unordered_map>
#include <set>
#include <fstream>
#include <chrono>
#include <random>
#include <bitset>
//#include <atcoder/all>
#define rep(i,n) for(int i=0;i<(n);i++)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) ((int)(x).size())
#define pb push_back
using ll = long long;
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
ll gcd(ll a, ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) {return a/gcd(a,b)*b;}
const ll mod = 1000000007;
ll mpow(ll a, ll x){
  ll res = 1;
  while(x > 0){
    if(x & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    x >>= 1;
  }
  return res;
}
vector<vector<ll>> mul(vector<vector<ll>> A, vector<vector<ll>> B){
  vector<vector<ll>> C(2,vector<ll>(2,0));
  rep(i,2) rep(j,2) rep(k,2){
    C[i][j] += A[i][k] * B[k][j] % (mod-1);
    C[i][j] %= (mod-1);
  }
  return C;
}
vector<vector<ll>> P(vector<vector<ll>> A,ll K){
  vector<vector<ll>> R = {{1,0},{0,1}};
  while(K>0){
    if(K&1) R = mul(R,A);
    A = mul(A,A);
    K>>=1;
  }
  return R;
}
int main(){
  const int max_N = 10000000;
  vector<int> prime(max_N+1, 0);
  prime[0] = prime[1] = 1;
  for(int i=2; i <= max_N; i++){
    if(prime[i]==0){
      for(int j=i*2; j <= max_N; j+=i){
        if(prime[j]==0) prime[j]=i;
      }
      prime[i]=i;
    }
  }
  ll N,K; cin >> N >> K;
  map<int,ll> F;
  while(N>1){
    int f = prime[N];
    F[f]++;
    N/=f;
  }
  while(1){
    map<int,ll> G;
    for(auto f:F){
      int n = f.first + 1;
      map<int,ll> tmp;
      while(n>1){
        int d = prime[n];
        tmp[d]++;
        n/=d;
      }
      for(auto t:tmp){
        G[t.first] += t.second * f.second % (mod-1);
        G[t.first] %= (mod-1);
      }
    }
    K--;
    swap(F,G);
    if(K==0) break;
    if(sz(F)==2 && F.count(2) && F.count(3)) break;
    if(sz(F)==1 && (F.count(2) || F.count(3))) break;
  }
  if(K){
    vector<vector<ll>> A = {{0,2},{1,0}};
    auto AK = P(A,K);
    map<int,ll> G;
    G[2] = (AK[0][0]*F[2]+AK[0][1]*F[3])%(mod-1);
    G[3] = (AK[1][0]*F[2]+AK[1][1]*F[3])%(mod-1);
    F = G;
  }
  ll ans = 1;
  for(auto f:F){
    ans *= mpow(f.first,f.second) % mod;
    ans %= mod;
  }
  cout << ans << '\n';
  return 0;
};
 
 
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