UVA 10870 - Recurrences 题目链接 题意:f(n) = a1 f(n - 1) + a2 f(n - 2) + a3 f(n - 3) + ... + ad f(n - d), for n > d. 已知前d项求第n项 思路:矩阵高速幂,相应矩阵为 |a1 a2 a3 ... ad| |1 0 0 ... 0 0 0| |0 1 0 ... 0 0 0| |0 0 1 ... 0 0 0| |0 0 0 ... 0 0 0| |0 0 0 ... 1 0 0| |0 0 0…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1811 矩阵快速幂 代码: #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<cmath> #include<set&…

题目传送门 题意:f(n) = a1f(n − 1) + a2f(n − 2) + a3f(n − 3) + . . . + adf(n − d), for n > d,求f (n) % m.训练指南的题目 分析:令:,.则 #include <bits/stdc++.h> int d, n, m; int a[16], f[16]; struct Mat { int m[17][17]; int row, col; Mat() { //row = col = 16; memset (m…

题意 求解递推式 \(f(n)=a_1*f(n-1)+a_2*f(n-2)+....+a_d*f(n-d)\) 的第 \(n\) 项模以 \(m\). \(1 \leq n \leq 2^{31}-1\) \(1 \leq m \leq 46340\) \(1 \leq d \leq 15\) 思路 矩阵乘法最经典的运用之一.先大致介绍一下矩阵乘法: 对于一个矩阵 \(A_{np}\) ,另一个矩阵 \(B_{pm}\) ,设它们的乘积为 \(C_{n,m}\) ,有 \(C_{i,j}=\di…

题意:给定 d , n , m (1<=d<=15,1<=n<=2^31-1,1<=m<=46340).a1 , a2 ..... ad.f(1), f(2) ..... f(d),求 f(n) = a1*f(n-1) + a2*f(n-2) +....+ ad*f(n-d),计算f(n) % m. 析:很明显的矩阵快速幂,构造矩阵, ,然后后面的就很简单了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1…

题目链接 https://odzkskevi.qnssl.com/d474b5dd1cebae1d617e6c48f5aca598?v=1524578553 题意 给出一个表达式 算法 f(n) 思路 n 很大 自然想到是 矩阵快速幂 那么问题就是 怎么构造矩阵 我们想到的一种构造方法是 n = 2 时 n = 3 时 然后大概就能够发现规律了吧 .. AC代码 #include <cstdio> #include <cstring> #include <ctype.h>…

给出一个d阶线性递推关系,求f(n) mod m的值. , 求出An-dv0,该向量的最后一个元素就是所求. #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; typedef long long Matrix[maxn][maxn]; typedef long long Vector[maxn]; int d, n, m; void matrix_mul(Mat…

题意:f(n) = a1f(n−1) + a2f(n−2) + a3f(n−3) + ... + adf(n−d), 计算这个f(n) 最重要的是推出矩阵. #include<cstdio> #include<cstring> #define ll long long ll mod, d, n; ll a[]; ll f[]; struct jz { ll num[][]; jz(){ memset(num, , sizeof(num)); } jz operator*(const…

Recurrences Input: standard input Output: standard output Consider recurrent functions of the following form: f(n) = a1 f(n - 1) + a2 f(n - 2) + a3 f(n - 3) + ... + ad f(n - d), for n > d. a1, a2, ..., ad - arbitrary constants. A famous example is th…

Problem ARecurrencesInput: standard inputOutput: standard output Consider recurrent functions of the following form: f(n) = a1 f(n - 1) + a2 f(n - 2) + a3 f(n - 3) + ... + ad f(n - d), for n > d.a1, a2, ..., ad - arbitrary constants. A famous example…