Yet another Number Sequence Let’s define another number sequence, given by the following function:f(0) = af(1) = bf(n) = f(n − 1) + f(n − 2), n > 1When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values…

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const int N = 4; int Mod; int msize; struct Mat { int mat[N][N]; }; Mat operator *(Mat a, Mat b) { Mat c; mems…

Description Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation: F1 = 1, F2 = 2, Fi = Fi - 1 + Fi - 2 (i > 2). We'll define a new number sequence Ai(k) by the formula: Ai(k) = Fi × ik (i ≥ 1). In thi…

Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multiple test cases. Each test case…

HDU - 1005 Number Sequence Problem Description A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.Given A, B, and n, you are to calculate the value of f(n). Input The input consists of multiple test…

A number sequence is defined as follows:  f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.  Given A, B, and n, you are to calculate the value of f(n). Input The input consists of multiple test cases. Each test case contains 3 integers…

Let’s define another number sequence, given by the following function: f(0) = a f(1) = b f(n) = f(n − 1) + f(n − 2), n > 1 When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values of a and b, you can get many different se…

题目:http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1607 题目描述 A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n). 输入…

题意:已知斐波那契数列fib(i) , 给你n 和 k , 求∑fib(i)*ik (1<=i<=n) 思路:不得不说,这道题很有意思,首先我们根据以往得出的一个经验,当我们遇到 X^k 的形式,当 X 很大,k很小时,我们可以利用二项式定理进行展开,然后求出递推式在利用矩阵加速 推导过程: 已知 fib(1) = 1, fib(2) = 1,fib(i) = fib(i-1) + fib(i-2); Ai(k) =fib(i)*i^k; 根据数学归纳法,我们可知 fib(i+1)*(i+1)…

题意: \(F_n\)为斐波那契数列,\(F_1=1,F_2=2\). 给定一个\(k\),定义数列\(A_i=F_i \cdot i^k\). 求\(A_1+A_2+ \cdots + A_n\). 分析: 构造一个列向量, \({\begin{bmatrix} F_{i-1}i^0 & F_{i-1}i^1 & \cdots & F_{i-1}i^k & F_{i}i^0 & F_{i}i^1 & \cdots & F_{i}i^k &…