#include<cstdio> #include<algorithm> #include<cstring> #include<iostream> using namespace std; ; struct Matrix { ][]; Matrix() { memset(a, , sizeof(a)); } Matrix operator * (const Matrix y) { Matrix ans; ; i <= ; i++) ; j <=…

E. Anniversary time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output There are less than 60 years left till the 900-th birthday anniversary of a famous Italian mathematician Leonardo Fibonacci. Of c…

题意:已知f(0) = a,f(1) = b,f(n) = f(n − 1) + f(n − 2), n > 1,求f(n)的后m位数. 分析:n最大为109,矩阵快速幂求解,复杂度log2(109). #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include&…

  Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13172   Accepted: 9368 Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci seque…

#include<bits/stdc++.h> #define mod 1000000009 using namespace std; typedef long long ll; typedef long long LL; struct Mat { LL mat[3][3]; Mat() { memset(mat,0,sizeof(mat)); } LL* operator [](int x) //注意这种写法 { return mat[x]; } } A; Mat Mut(Mat a,Mat…

斐波那契额数列的第N项 斐波那契数列的定义如下: F(0) = 0 F(1) = 1 F(n) = F(n - 1) + F(n - 2) (n >= 2) (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...) 给出n,求F(n),由于结果很大,输出F(n) % 1000000009的结果即可. 输入 输入1个数n(1 <= n <= 10^18). 输出 输出F(n) % 1000000009的结果. 输入样例 11 输出…

题目链接: http://codeforces.com/gym/101161/attachments 题意: $T$组数据 每组数据包含$L,R,K$ 计算$\sum_{k|n}^{}F(n)$ 定义$F(n)$为斐波那契数列第$n$项 数据范围: $1\leq T\leq 10000$ $1\leq L\leq 10^{18}$ $1\leq R\leq 10^{18}$ 分析: 博客来源:https://blog.csdn.net/qq_41552508/article/details/97…

先占坑 后面再写详细的 import numpy as np def pow(n): a = np.array([[1,0],[0,1]]) b = np.array([[1,1],[1,0]]) n -= 1 while(n > 0): if (n % 2 == 1): a = np.dot(b, a) b = np.dot(b, b) n >>= 1 return a[0][0] n = int(input()) print(factorial(n))…

这题并不复杂. 设$A=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ 由题中公式: $\begin{pmatrix}f(n+1) & f(n)\\ f(n+1) & f(n-1)\end{pmatrix} = {\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}}^{n}$ 可知,若要求f(n)只要求矩阵A的n次方即可.设我们所需的矩阵为$Answer$. 对于此题,我们可以先将…

典型的两道矩阵快速幂求斐波那契数列 POJ 那是 默认a=0,b=1 UVA 一般情况是 斐波那契f(n)=(n-1)次幂情况下的(ans.m[0][0] * b + ans.m[0][1] * a): //POJ #include <cstdio> #include <iostream> using namespace std; ; struct matrix { ][]; }ans, base; matrix multi(matrix a, matrix b) { matrix…