結果
問題 | No.891 隣接3項間の漸化式 |
ユーザー |
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提出日時 | 2020-08-12 18:30:45 |
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 | 3,024 bytes |
コンパイル時間 | 1,482 ms |
コンパイル使用メモリ | 113,496 KB |
実行使用メモリ | 6,824 KB |
最終ジャッジ日時 | 2024-10-09 18:32:04 |
合計ジャッジ時間 | 2,879 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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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 <string>#include <vector>#include <queue>#include <deque>#include <algorithm>#include <set>#include <map>#include <bitset>#include <cmath>#include <functional>#include <iomanip>#define vll vector<ll>#define vvvl vector<vvl>#define vvl vector<vector<ll>>#define VV(a, b, c, d) vector<vector<d>>(a, vector<d>(b, c))#define VVV(a, b, c, d) vector<vvl>(a, vvl(b, vll (c, d)));#define re(c, b) for(ll c=0;c<b;c++)#define all(obj) (obj).begin(), (obj).end()typedef long long int ll;typedef long double ld;using namespace std;// matrix<集合> (集合, 加法における単位元, 加算, 乗算)template<typename T>struct matrix{function<T(T, T)> f, g;vector<vector<T>> mat;T unit;matrix(vector<vector<T>> v, T UNIT, function<T(T, T)> add_func,function<T(T, T)> multi_func):f(add_func), g(multi_func), unit(UNIT){mat = v;}matrix<T> t(){vector<vector<T>> ret(mat[0].size(), vector<T>(mat.size()));re(i, mat[0].size()) re(j, mat.size()) ret[i][j] = mat[j][i];return matrix<T>(ret, f, g);}matrix<T> operator * (matrix<T> B){vector<vector<T>> ret(mat.size(), vector<T>(B.mat[0].size(), unit));if(mat[0].size()!=B.mat.size()){std::cout << "matrix operator " << "*" << " error" << '\n';return matrix<T>(ret, unit, f, g);}re(i, mat.size()){re(j, B.mat[0].size()){re(k, mat[0].size()){ret[i][j] = f(ret[i][j], g(mat[i][k], B.mat[k][j]));}}}return matrix<T>(ret, unit, f, g);}matrix<T> operator * (ll num){vector<vector<T>> ret = mat;re(i, mat.size()) re(j, mat[0].size()) ret[i][j] *= num;return matrix<T>(ret, unit, f, g);}matrix<T> operator + (matrix<T> B){vector<vector<T>> ret = mat;if(mat.size()!=B.mat.size()||mat[0].size()!=B.mat[0].size()){std::cout << "matrix operator " << "+" << " error" << '\n';return matrix<T>(ret, unit, f, g);}re(i, mat.size()) re(j, mat[0].size()) ret[i][j] = f(mat[i][j], B.mat[i][j]);return matrix<T>(ret, unit, f, g);}matrix<T> operator ^ (ll num){matrix<T> ret(vector<vector<T>> (0, vector<T>(0)), unit, f, g);if(mat.size()!=mat[0].size()){std::cout << "matrix operator " << "^" << " error" << '\n';return ret;}matrix<T> tmp(mat, unit, f, g);bool flag = false;while(num>0){if(num%2){if(flag) ret = ret * tmp;else ret.mat = tmp.mat, flag = true;}num/=2, tmp = tmp * tmp;}return ret;}vector<T> & operator [](int n){return mat[n];}};#define P 1000000007ll ad(ll a, ll b){return (a+b)%P;}ll ml(ll a, ll b){return (a*b)%P;}int main(int argc, char const *argv[]) {ll a, b, n;std::cin >> a >> b >> n;matrix<ll> A({{1},{0}}, 0, ad, ml);matrix<ll> B({{a, b}, {1, 0}}, 0, ad, ml);if(n==0) std::cout << 0 << '\n';if(n==1) std::cout << 1 << '\n';if(n>1) std::cout << ((B^(n-1)) * A)[0][0] << '\n';return 0;}