結果

問題 No.8037 Restricted Lucas (Hard)
ユーザー okaduki
提出日時 2018-04-02 00:25:27
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 1,920 bytes
コンパイル時間 926 ms
コンパイル使用メモリ 83,032 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-26 06:22:30
合計ジャッジ時間 1,479 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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 * 1
other AC * 6
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 <iostream>
#include <vector>
#include <cmath>
#include <functional>
using namespace std;
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using LL = long long;
using VL = vector<LL>;
using VVL = vector<VL>;
using PLL = pair<LL, LL>;
using VS = vector<string>;
#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))
LL MOD;
int zero;
LL one;
LL two;
LL three;
LL four;
LL five;
LL six;
LL seven;
LL eight;
LL nine;
LL ten;
#define FOR(i,a,b) for(LL i=(a);i<(b);i=plus<LL>()(i,one))
#define REP(i,n)  FOR(i,zero,n)
LL mod(LL a){
  return div(a, MOD).rem;
}
LL mul(LL a, LL b){
  return mod(multiplies<LL>()(a,b));
}
VVL mul(const VVL& a, const VVL& b){
  VVL tmp(two, VL(two));
  REP(i,two) REP(j,two) REP(k,two){
    tmp[i][j] = plus<LL>()(tmp[i][j], mul(a[i][k], b[k][j]));
    tmp[i][j] = mod(tmp[i][j]);
  }
  return tmp;
}
LL solve(LL n){
  VVL xs = {{one,one},{one,zero}};
  VVL p = {{one,zero},{zero,one}};
  while(n > zero){
    if(n&one){
      p = mul(p, xs);
    }
    xs = mul(xs, xs);
    n >>= one;
  }
  return mod(plus<LL>()(p[one][zero],(p[one][one]<<one)));
}
int main(){
  one = !!M_PI;
  two = plus<LL>()(one, one);
  three = plus<LL>()(two, one);
  four = plus<LL>()(three, one);
  five = plus<LL>()(four, one);
  six = plus<LL>()(five, one);
  seven = plus<LL>()(six, one);
  eight = plus<LL>()(seven, one);
  nine = plus<LL>()(eight, one);
  ten = plus<LL>()(nine, one);
  MOD = ten;
  MOD = multiplies<LL>()(MOD, MOD);
  MOD = multiplies<LL>()(MOD, MOD);
  MOD = multiplies<LL>()(MOD, MOD);
  MOD = plus<LL>()(multiplies<LL>()(MOD, ten), seven);
  
  int T;
  cin >> T;
  REP(t,T){
    LL n;
    cin >> n;
    cout << solve(n) << endl;
  }
}
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