This documentation is automatically generated by online-judge-tools/verification-helper
#include "src/fps/bostan_mori.hpp"
mint bostan_mori(FPS<mint> a, FPS<mint> c, ll k)
斉次線形漸化式
\[a_i = \sum\limits_{j = 0}^{d - 1} c_j a_{i - 1 - j}\]の $a$ の前 $n ~(\geq d)$ 項と $c$ が与えられたときに $a_k$ を返します.
制約
計算量
#pragma once
#include "../template/template.hpp"
template <template <typename> typename FPS, typename mint>
mint bostan_mori(const FPS<mint>& a, const FPS<mint>& c, long long k) {
assert(k >= 0);
if(k < (int)a.size()) return a[k];
assert(a.size() >= c.size());
FPS<mint> q = FPS<mint>{1} - (c << 1);
FPS<mint> p = (a * q).pre((int)c.size());
if(p.empty()) return 0;
while(k > 0) {
auto q2 = q;
for(int i = 1; i < (int)q2.size(); i += 2) q2[i] = -q2[i];
const auto s = p * q2;
const auto t = q * q2;
if(k & 1) {
for(int i = 1; i < (int)s.size(); i += 2) p[i >> 1] = s[i];
for(int i = 0; i < (int)t.size(); i += 2) q[i >> 1] = t[i];
} else {
for(int i = 0; i < (int)s.size(); i += 2) p[i >> 1] = s[i];
for(int i = 0; i < (int)t.size(); i += 2) q[i >> 1] = t[i];
}
k >>= 1;
}
return p[0];
}
#line 2 "src/template/template.hpp"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<long long, long long>;
#define rep(i, a, b) for(long long i = (a); i < (b); ++i)
#define rrep(i, a, b) for(long long i = (a); i >= (b); --i)
constexpr long long inf = 4e18;
struct SetupIO {
SetupIO() {
ios::sync_with_stdio(0);
cin.tie(0);
cout << fixed << setprecision(30);
}
} setup_io;
#line 3 "src/fps/bostan_mori.hpp"
template <template <typename> typename FPS, typename mint>
mint bostan_mori(const FPS<mint>& a, const FPS<mint>& c, long long k) {
assert(k >= 0);
if(k < (int)a.size()) return a[k];
assert(a.size() >= c.size());
FPS<mint> q = FPS<mint>{1} - (c << 1);
FPS<mint> p = (a * q).pre((int)c.size());
if(p.empty()) return 0;
while(k > 0) {
auto q2 = q;
for(int i = 1; i < (int)q2.size(); i += 2) q2[i] = -q2[i];
const auto s = p * q2;
const auto t = q * q2;
if(k & 1) {
for(int i = 1; i < (int)s.size(); i += 2) p[i >> 1] = s[i];
for(int i = 0; i < (int)t.size(); i += 2) q[i >> 1] = t[i];
} else {
for(int i = 0; i < (int)s.size(); i += 2) p[i >> 1] = s[i];
for(int i = 0; i < (int)t.size(); i += 2) q[i >> 1] = t[i];
}
k >>= 1;
}
return p[0];
}