杜教BM模板

用于求线性递推式第n项，扔进前k+x项即可

Code

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
typedef long long ll;
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<ll> VI;
typedef pair<ll, ll> PII;
const ll mod = 10000;
ll powmod(ll a, ll b) {
  ll res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
ll _, n;
namespace linear_seq {
const ll N = 10010;
ll res[N], base[N], _c[N], _md[N];
vector<ll> Md;
void mul(ll *a, ll *b, int k) {
  rep(i, 0, k + k) _c[i] = 0;
  rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] =
      (_c[i + j] + a[i] * b[j]) % mod;
  for (ll i = k + k - 1; i >= k; i--)
    if (_c[i])
      rep(j, 0, SZ(Md)) _c[i - k + Md[j]] =
          (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
  rep(i, 0, k) a[i] = _c[i];
}
ll solve(ll n, VI a, VI b) {
  ll ans = 0, pnt = 0;
  ll k = SZ(a);
  assert(SZ(a) == SZ(b));
  rep(i, 0, k) _md[k - 1 - i] = -a[i];
  _md[k] = 1;
  Md.clear();
  rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);
  rep(i, 0, k) res[i] = base[i] = 0;
  res[0] = 1;
  while ((1ll << pnt) <= n) pnt++;
  for (ll p = pnt; p >= 0; p--) {
    mul(res, res, k);
    if ((n >> p) & 1) {
      for (ll i = k - 1; i >= 0; i--) res[i + 1] = res[i];
      res[0] = 0;
      rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
    }
  }
  rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
  if (ans < 0) ans += mod;
  return ans;
}
VI BM(VI s) {
  VI C(1, 1), B(1, 1);
  ll L = 0, m = 1, b = 1;
  rep(n, 0, SZ(s)) {
    ll d = 0;
    rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
    if (d == 0)
      ++m;
    else if (2 * L <= n) {
      VI T = C;
      ll c = mod - d * powmod(b, mod - 2) % mod;
      while (SZ(C) < SZ(B) + m) C.pb(0);
      rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
      L = n + 1 - L;
      B = T;
      b = d;
      m = 1;
    } else {
      ll c = mod - d * powmod(b, mod - 2) % mod;
      while (SZ(C) < SZ(B) + m) C.pb(0);
      rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
      ++m;
    }
  }
  return C;
}
ll gao(VI a, ll n) {
  VI c = BM(a);
  c.erase(c.begin());
  rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
  return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
}
};  // namespace linear_seq
int main() {
  while (~scanf("%lld", &n) && n != -1) {
    vector<ll>v;
       v.push_back(0);
       v.push_back(1);
        v.push_back(1);
        v.push_back(2);
        v.push_back(3);
        v.push_back(5);
        v.push_back(8);
        v.push_back(13);
        v.push_back(21);
        v.push_back(34);
      printf("%lld\n", linear_seq::gao(v, n));
  }
  return 0;
}