結果
問題 | No.891 隣接3項間の漸化式 |
ユーザー |
|
提出日時 | 2021-02-04 22:17:19 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 9 ms / 2,000 ms |
コード長 | 3,295 bytes |
コンパイル時間 | 1,263 ms |
コンパイル使用メモリ | 136,292 KB |
最終ジャッジ日時 | 2025-01-18 11:17:07 |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 39 |
ソースコード
#include <iostream>#include <vector>#include <algorithm>#include <cmath>#include <queue>#include <string>#include <map>#include <set>#include <stack>#include <tuple>#include <deque>#include <array>#include <numeric>#include <bitset>#include <iomanip>#include <cassert>#include <chrono>#include <random>#include <limits>#include <iterator>#include <functional>#include <sstream>#include <fstream>#include <complex>#include <cstring>#include <unordered_map>#include <unordered_set>using namespace std;using ll = long long;constexpr int INF = 1001001001;constexpr int mod = 1000000007;// constexpr int mod = 998244353;template<class T>inline bool chmax(T& x, T y){if(x < y){x = y;return true;}return false;}template<class T>inline bool chmin(T& x, T y){if(x > y){x = y;return true;}return false;}struct mint {int x;mint() : x(0) {}mint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}mint& operator+=(const mint& p){if((x += p.x) >= mod) x -= mod;return *this;}mint& operator-=(const mint& p){if((x -= p.x) < 0) x += mod;return *this;}mint& operator*=(const mint& p){x = (int)(1LL * x * p.x % mod);return *this;}mint& operator/=(const mint& p){*this *= p.inverse();return *this;}mint operator-() const { return mint(-x); }mint operator+(const mint& p) const { return mint(*this) += p; }mint operator-(const mint& p) const { return mint(*this) -= p; }mint operator*(const mint& p) const { return mint(*this) *= p; }mint operator/(const mint& p) const { return mint(*this) /= p; }bool operator==(const mint& p) const { return x == p.x; }bool operator!=(const mint& p) const { return x != p.x; }mint pow(int64_t n) const {mint res = 1, mul = x;while(n > 0){if(n & 1) res *= mul;mul *= mul;n >>= 1;}return res;}mint inverse() const { return pow(mod - 2); }friend ostream& operator<<(ostream& os, const mint& p){return os << p.x;}friend istream& operator>>(istream& is, mint& p){int64_t val;is >> val;p = mint(val);return is;}};using Vec = vector<mint>;using Mat = vector<Vec>;Mat Mul(const Mat& A, const Mat& B){Mat C(A.size(), Vec(B[0].size()));for(int i = 0; i < (int)A.size(); ++i){for(int k = 0; k < (int)A[0].size(); ++k){for(int j = 0; j < (int)B[0].size(); ++j){C[i][j] += A[i][k] * B[k][j];}}}return C;}Mat Pow(Mat A, int n){Mat B(A.size(), Vec(A.size()));for(int i = 0; i < (int)A.size(); ++i) B[i][i] = 1;while(n > 0){if(n & 1) B = Mul(B, A);A = Mul(A, A);n >>= 1;}return B;}int main(){ios::sync_with_stdio(false);cin.tie(nullptr);int a, b, n;cin >> a >> b >> n;if(n <= 1){cout << n << endl;return 0;}Mat W(2, Vec(2));W[0][0] = a; W[0][1] = b;W[1][0] = 1; W[1][1] = 0;W = Pow(W, n - 1);cout << W[0][0] << endl;return 0;}