任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6395 Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2564    Accepted Submission(s): 999 Problem Description Let us define a sequence as…

Problem Description Holion August will eat every thing he has found. Now there are many foods,but he does not want to eat all of them at once,so he find a sequence. fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise He gives you 5 numbers n,a,b,c,p,and he will…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5667 题意: Lcomyn 是个很厉害的选手,除了喜欢写17kb+的代码题,偶尔还会写数学题.他找到了一个数列: fn= 1,ab,abfcn−1fn−2,n=1n=2otherwise 给定各个数,求fn. 分析: 可以发现最后都是a的倍数,这样我们让fn对a取对数,令tn=logafn方程就转化为b+ctn−1+tn−2,这样利用矩阵快速幂直接算幂数,最后快速幂一下就可以了. 注意: 由费马小…

题意: 已知\(A,B,C,D,P,n\)以及 \[\left\{ \begin{aligned} & F_1 = A \\ & F_2 = B\\ & F_n = C*F_{n-2} + D*F_{n-2}+\lfloor(\frac{P}{n})\rfloor \end{aligned} \right. \] ,求\(F_n \ mod\ (1e9e+7)\),\(n \leq 1e9\) 思路: 显然\(\lfloor(\frac{P}{n})\rfloor\)相同的情况是有…

http://acm.hdu.edu.cn/showproblem.php?pid=6395 Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1475    Accepted Submission(s): 539 Problem Description Let us define a sequence as belo…

实际上,对于位数相同的连续段,可以用矩阵快速幂求出最后的ans,那么题目中一共只有18个连续段. 分段矩阵快速幂即可. #include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<queue> #include<cmath> #define ll long long using na…

HDU.1575 Tr A ( 矩阵快速幂) 点我挑战题目 题意分析 直接求矩阵A^K的结果,然后计算正对角线,即左上到右下对角线的和,结果模9973后输出即可. 由于此题矩阵直接给出的,题目比较裸. 代码总览 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <sstream> #include <set> #…

斐波那契数列后四位可以用快速幂取模(模10000)算出.前四位要用公式推 HDU 3117 Fibonacci Numbers(矩阵快速幂+公式) f(n)=(((1+√5)/2)^n+((1-√5)/2)^n)/√5 假设F[n]可以表示成 t * 10^k(t是一个小数),那么对于F[n]取对数log10,答案就为log10 t + K,此时很明显log10 t<1,于是我们去除整数部分,就得到了log10 t 再用pow(10,log10 t)我们就还原回了t.将t×1000就得到了F[n…

Your job is simple, for each task, you should output Fn module 109+7. Input The first line has only one integer T, indicates the number of tasks. Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n. 1≤T≤200≤A,B,C,D≤1091≤P,…

思路:一开始不会n^4的推导,原来是要找n和n-1的关系,这道题的MOD是long long 的,矩阵具体如下所示 最近自己总是很坑啊,代码都瞎吉坝写,一个long long的输入写成%d一直判我TLE,一度怀疑矩阵快速幂地复杂度orz 代码: #include<set> #include<cstring> #include<cstdio> #include<algorithm> #define ll long long const int maxn = 7…