結果

問題 No.194 フィボナッチ数列の理解(1)
ユーザー SHIJOU
提出日時 2020-08-12 12:51:22
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,730 bytes
コンパイル時間 3,187 ms
コンパイル使用メモリ 207,284 KB
最終ジャッジ日時 2025-01-12 21:18:03
ジャッジサーバーID
（参考情報） 		judge2 / judge2
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ファイルパターン 結果
sample AC * 3
other AC * 29 WA * 8
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ソースコード

diff #
プレゼンテーションモードにする

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 100010001;
const ll LINF = (ll)INF*INF*10;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}
const int mod = 1000000007;
//const int mod = 998244353;
struct mint {
  int64_t x;
  mint(int64_t x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(int64_t t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
  //for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) {return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
//head
int n;
ll k;
using V = vector<mint>;
using M = vector<V>;
M mul(M &a, M b) {
  int n = a.size();
  M res(n, V(n));
  rep(i, n) rep(j, n) rep(k, n) res[i][j] += a[i][k] * b[k][j];
  return res;
}
M pow(M &a, ll p) {
  int n = a.size();
  M res(n, V(n));
  if(!p) {
    rep(i, n) res[i][i] = 1;
    return res;
  }
  M ano = pow(a, p>>1);
  res = mul(ano, ano);
  if(p&1) res = mul(res, a);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  if(n > 30) {
    vi a(n);
    rep(i, n) cin >> a[i];
    if(k <= n) {
      cout << a[k-1] << ' ' << accumulate(a.begin(), a.begin() + k, mint()) << endl;
    }
    else {
      mint now = a[n-1], sum1, sum2 = accumulate(all(a), mint());
      queue<mint> q;
      rep(i, n) q.emplace(a[i]);
      rep(i, n-k) {
        now = sum2-sum1;
        sum2 += now;
        sum1 += q.front();
        q.pop();
        q.emplace(now);
      }
      cout << now << ' ' << sum2 << endl;
    }
  }
  else {
    V a(n);
    rep(i, n) cin >> a[i];
    if(k <= n) {
      cout << a[k-1] << ' ' << accumulate(a.begin(), a.begin()+k, mint()) << endl;
    }
    else {
      M mat(n+1, V(n+1));
      rep(i, n+1) mat[0][i] = 1;
      rep(i, n) mat[1][i+1] = 1;
      rep(i, n-1) mat[i+2][i+1] = 1;
      //rep(i, n+1) rep(j, n+1) cout << mat[i][j] << (j == n?'\n':' ');
      mat = pow(mat, k-n);
      //rep(i, n+1) rep(j, n+1) cout << mat[i][j] << (j == n?'\n':' ');
      mint s = accumulate(all(a), mint());
      mint ans;
      rep(i, n) ans += a[n-1-i] * mat[1][i+1];
      mint sum;
      sum += s*mat[0][0];
      rep(i, n) sum += a[n-1-i] * mat[0][i+1];
      cout << ans << ' ' << sum << endl;
    }
  }
}
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