用矩阵求解线性递推式通项 用fft优化矩阵乘法 首先把递推式求解转化为矩阵求幂,再利用特征多项式f(λ)满足f(A) = 0,将矩阵求幂转化为多项式相乘, 最后利用傅里叶变换的高效算法(迭代取代递归)(参见算法导论)解决. #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <string> #include <vector>…

题解见X姐的论文 矩阵乘法递推的优化.仅仅是mark一下. .…

Recursive sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn,…

Recursive sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 249    Accepted Submission(s): 140 Problem Description Farmer John likes to play mathematics games with his N cows. Recently, t…

题目链接:CF1106F Lunar New Year and a Recursive Sequence 大意:已知\(f_1,f_2,\cdots,f_{k-1}\)和\(b_1,b_2,\cdots,b_k\),且有递推关系 \[ f_i=(\prod_{j=1}^kf_{i-j}^{b_j})\text%p \] 对于所有\(i>k\)均成立,给出\(f_n=m\),求\(f_k\),\(p=998244353\) 分析: 数论板子大集合 首先很显然有\(f_n=f_k^{q}\),于是考…

题目链接:Recursive sequence 题意:给出前两项和递推式,求第n项的值. 题解:递推式为:$F[i]=F[i-1]+2*f[i-2]+i^4$ 主要问题是$i^4$处理,容易想到用矩阵快速幂,那么$i^4$就需要从$(i-1)$转移过来. $ i^4 = (i-1)^4 + 4*(i-1)^3 + 6*(i-1)^2 + 4*(i-1) + 1$ $f_i$ $f_{i-1}$ $i^4$ $i^3$ $i^2$ $i$ $1$ = $f_{i-1}$ $f_{i-2}$ $(i…

Recursive sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3832    Accepted Submission(s): 1662 Problem Description Farmer John likes to play mathematics games with his N cows. Recently…

我诈尸啦! 高三退役选手好不容易抛弃天利和金考卷打场CF,结果打得和shi一样--还因为queue太长而unrated了!一个学期不敲代码实在是忘干净了-- 没分该没分,考题还是要订正的 =v= 欢迎阅读本题解! P.S. 这几个算法我是一个也想不起来了 TAT 题目链接 Codeforces 1106F Lunar New Year and a Recursive Sequence 新年和递推数列 题意描述 某数列\(\{f_i\}\)递推公式:\[f_i = (\prod_{j=1}^kf_…

题目链接:传送门 题目: Recursive sequence Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Farmer John likes to play mathematics games with his N cows. Recently, they are attracted…

题目传送门 题目描述:给出一个数列的第一项和第二项,计算第n项. 递推式是 f(n)=f(n-1)+2*f(n-2)+n^4. 由于n很大,所以肯定是矩阵快速幂的题目,但是矩阵快速幂只能解决线性的问题,n^4在这个式子中是非线性的,后一项和前一项没有什么直接关系,所以模拟赛的时候想破头也不会做. 这里要做一个转换,把n^4变成一个线性的,也就是和(n-1)^4有关系的东西,而这个办法就是: n^4=(n-1+1)^4=(n-1)^4+4*(n-1)^3+6*(n-1)^2+4*(n-1)^1+(…