結果

問題 No.492 IOI数列
コンテスト
ユーザー koyumeishi
提出日時 2017-03-11 01:13:49
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 1,000 ms
コード長 7,206 bytes
コンパイル時間 1,422 ms
コンパイル使用メモリ 124,256 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-24 09:26:48
合計ジャッジ時間 2,220 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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 * 19
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 <iostream>
#include <vector>
#include <cstdio>
#include <sstream>
#include <map>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <functional>
#include <set>
#include <ctime>
#include <random>
#include <chrono>
#include <cassert>
#include <tuple>
#include <utility>
using namespace std;




namespace {
  using Integer = long long; //__int128;
  template<class T, class S> istream& operator >> (istream& is, pair<T,S>& p){return is >> p.first >> p.second;}
  template<class T> istream& operator >> (istream& is, vector<T>& vec){for(T& val: vec) is >> val; return is;}
  template<class T> istream& operator ,  (istream& is, T& val){ return is >> val;}
  template<class T, class S> ostream& operator << (ostream& os, const pair<T,S>& p){return os << p.first << " " << p.second;}
  template<class T> ostream& operator << (ostream& os, const vector<T>& vec){for(size_t i=0; i<vec.size(); i++) os << vec[i] << (i==vec.size()-1?"":" "); return os;}
  template<class T> ostream& operator ,  (ostream& os, const T& val){ return os << " " << val;}

  istream& operator >> (istream& is, __int128& x){
  static char buff[100];
  is >> buff;
  x = 0;
  size_t i = 0;
  if(buff[0] == '-') i++;
  for(; buff[i]; i++){
    x *= 10;
    x += buff[i]-'0';
  }
  if(buff[0] == '-') x *= -1;
  return is;
}

ostream& operator << (ostream& os, __int128 x){
  if(x==0) return os << 0;
  string s;
  if(x<0){
    os << "-";
    x *= -1;
  }
  while(x!=0){
    s += (x%10) + '0';
    x /= 10;
  }
  reverse(s.begin(), s.end());
  return os << s;
}

  template<class H> void print(const H& head){ cout << head; }
  template<class H, class ... T> void print(const H& head, const T& ... tail){ cout << head << " "; print(tail...); }
  template<class ... T> void println(const T& ... values){ print(values...); cout << endl; }

  template<class H> void eprint(const H& head){ cerr << head; }
  template<class H, class ... T> void eprint(const H& head, const T& ... tail){ cerr << head << " "; eprint(tail...); }
  template<class ... T> void eprintln(const T& ... values){ eprint(values...); cerr << endl; }

  class range{ Integer start_, end_, step_; public: struct range_iterator{ Integer val, step_; range_iterator(Integer v, Integer step) : val(v), step_(step) {} Integer operator * (){return val;} void operator ++ (){val += step_;} bool operator != (range_iterator& x){return step_ > 0 ? val < x.val : val > x.val;} }; range(Integer len) : start_(0), end_(len), step_(1) {} range(Integer start, Integer end) : start_(start), end_(end), step_(1) {} range(Integer start, Integer end, Integer step) : start_(start), end_(end), step_(step) {} range_iterator begin(){ return range_iterator(start_, step_); } range_iterator   end(){ return range_iterator(  end_, step_); } };

  inline string operator "" _s (const char* str, size_t size){ return move(string(str)); }
  constexpr Integer my_pow(Integer x, Integer k, Integer z=1){return k==0 ? z : k==1 ? z*x : (k&1) ? my_pow(x*x,k>>1,z*x) : my_pow(x*x,k>>1,z);}
  constexpr Integer my_pow_mod(Integer x, Integer k, Integer M, Integer z=1){return k==0 ? z%M : k==1 ? z*x%M : (k&1) ? my_pow_mod(x*x%M,k>>1,M,z*x%M) : my_pow_mod(x*x%M,k>>1,M,z);}
  constexpr unsigned long long operator "" _ten (unsigned long long value){ return my_pow(10,value); }

  inline int k_bit(Integer x, int k){return (x>>k)&1;} //0-indexed

  mt19937 mt(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count());

  template<class T> string join(const vector<T>& v, const string& sep){ stringstream ss; for(size_t i=0; i<v.size(); i++){ if(i>0) ss << sep; ss << v[i]; } return ss.str(); }

  inline string operator * (string s, int k){ string ret; while(k){ if(k&1) ret += s; s += s; k >>= 1; } return ret; }
}
constexpr long long mod = 9_ten + 7;

long long gcd(long long a, long long b){ return (b==0)?a:gcd(b,a%b); }
template<class ... T> long long gcd(long long a, long long b, T ... c){ return gcd(gcd(a,b), c...);}
long long lcm(long long a, long long b){ if(a<b) swap(a,b); return (b==1)?a:a*(b/gcd(a,b)); }
template<class ... T> long long lcm(long long a, long long b, T ... c){ return lcm(lcm(a,b), c...);}

long long extgcd(long long a, long long b, long long &x, long long &y){
  long long d=a;
  if(b!=0){
    d = extgcd(b, a%b, y, x);
    y -= (a/b) * x;
  }else{
    x = 1;
    y = 0;
  }
  return d;
}


long long mod_inverse(long long a, long long m){
  long long x,y;
  extgcd(a,m,x,y);
  return (m+x%m)%m;
}


template<class T>
vector< vector<T> > multmat(const vector<vector<T> > &A, const vector<vector<T>> &B, int n, int p, int m, T mod){
  vector<vector<T> > C(n, vector<T>(m,0));
  for(int i=0; i<n; i++){
    for(int k=0; k<p; k++){
      for(int j=0; j<m; j++){
        C[i][j] += A[i][k] * B[k][j];
        C[i][j] %= mod;
      }
    }
  }
  return C;
}

//A[n*n]^k 
template<class T>
vector< vector<T> > mat_pow(vector<vector<T> > A, long long k, T mod){
  int n = A.size();
  vector<vector<T> > ret(n, vector<T>(n, 0) );
  for(int i=0; i<n; i++){
    ret[i][i] = 1;
  }
  while(k>0){
    if(k&1) ret = multmat(A,ret, n,n,n, mod);
    A = multmat(A,A, n,n,n, mod);
    k>>=1;
  }
  return ret;
}

template<class T>
T Chinese_Remainder_Theorem_Garner(vector<long long> x, vector<long long> y, T MOD){
  int N = x.size();
  bool valid = true;
  //前処理
  //gcd(Yi,Yj) == 1 (i!=j) でなくてはならないので、
  //共通の因数 g = gcd(Yi,Yj) を見つけたら片側に寄せてしまう
  for(int i=0; i<N; i++){
    for(int j=i+1; j<N; j++){
      if(i == j) continue;
      long long g = gcd(y[i], y[j]);

      if( x[i]%g != x[j]%g ) valid = false; //解が存在しない

      if(g != 1){
        y[i] /= g; y[j] /= g;
        long long g_ = gcd(y[i], g);
        while(g_ != 1){
          y[i] *= g_;
          g /= g_;
          g_ = gcd(y[i], g);
        }
        y[j] *= g;

        x[i] %= y[i];
        x[j] %= y[j];
      }
    }
  }

  if(!valid){
    cerr << -1 << endl;
    return 0;
  }

  //Garner's algorithm
  vector<T> z(N);
  for(int i=0; i<N; i++){
    z[i] = x[i];
    for(int j=0; j<i; j++){
      z[i] = mod_inverse(y[j], y[i]) % y[i] * (z[i] - z[j]) % y[i];

      z[i] = (z[i]+y[i])%y[i];
    }
  }

  __int128 ans = 0;
  __int128 tmp = 1;
  for(int i=0; i<N; i++){
    ans = (ans + z[i] * tmp)%MOD;
    tmp = (tmp * y[i])%MOD;
  }

  return ans;
}

int main(){
  long long n;
  cin >> n;

  if(n==1){
    println(1);
    println(1);
    return 0;
  }

  vector<vector<long long>> A = {
    {100, 1},
    {  0, 1}
  };
  auto AA = A;

  vector<long long> M = {
    11,
23,
4093,
8779,
21649,
513239,
mod
  };


  vector<vector<vector<long long>>> B(M.size());
  for(int i=0; i<M.size(); i++){
    A = AA;
    for(auto& v : A) for(auto& x : v) x %= M[i];
    B[i] = mat_pow(A, n-1, M[i]);
  }

  vector<long long> C(M.size());
  for(int i=0; i<M.size(); i++){
    C[i] = (B[i][0][0] + B[i][0][1]) % M[i];
  }

  println( C.back() );
  C.pop_back();
  M.pop_back();

  stringstream ss;
  ss << "101010101010101010101";
  __int128 MOOOD;
  ss >> MOOOD;
  __int128 c = Chinese_Remainder_Theorem_Garner<__int128>(C, M, MOOOD);
  println( c );

  return 0;
}
0