题目链接 :http://acm.hdu.edu.cn/showproblem.php?pid=6030 Problem Description Little Q wants to buy a necklace for his girlfriend. Necklaces are single strings composed of multiple red and blue beads. Little Q desperately wants to impress his girlfriend,…

题目链接 Problem Description Function Fx,ysatisfies: For given integers N and M,calculate Fm,1 modulo 1e9+7. Input There is one integer T in the first line. The next T lines,each line includes two integers N and M . 1<=T<=10000,1<=N,M<2^63. Output…

Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 187893    Accepted Submission(s): 46820 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A…

题意:已知F0 = 0,F1 = 1,Fn = Fn - 1 + Fn - 2(n >= 2), 且若n=Fa1+Fa2+...+Fak where 0≤a1≤a2≤⋯≤a,n为正数,则n为mjf-good,否则为mjf-bad,给定k,求最小的mjf-bad. 分析:找规律可得,F2*k+3 - 1,矩阵快速幂即可. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype&g…

Happy Necklace 前天个人赛规律都找出来了,n的范围是\(10^{18}\),我一想GG,肯定是矩阵快速幂,然后就放弃了. 昨天学了一下矩阵快速幂. 题意 现在小Q要为他的女朋友一个有n个宝石的好的项链(直线),定义好的项链为:如果任意素数长的子串中蓝宝石的数量不小于红宝石的数量就是好的.小Q可以买多少种好的项链呢? 思路 n这么大,绝对是规律题. 先简单分析一下,其实对于好项链的定义可以化简为任意长度为3的子串中至少有两个蓝宝石,然后打表. #include<bits/stdc++…

http://acm.hdu.edu.cn/showproblem.php?pid=2855 化简这个公式,多写出几组就会发现规律 d[n]=F[2*n] 后面的任务就是矩阵快速幂拍一个斐波那契模板出来了 这里用的是2维 vector #include<iostream> #include<cstdio> #include<vector> using namespace std; typedef vector<int>vec; typedef vector&…

http://acm.hdu.edu.cn/showproblem.php?pid=5950 题意:给出 a,b,n,递推出 f(n) = f(n-1) + f(n-2) * 2 + n ^ 4. f(1) = a, f(2) = b. 思路:在比赛时候知道是矩阵快速幂,可是推不出矩阵.那个n^4不知道怎么解决.结束后问其他人才知道要构造一个7 * 7的矩阵,而不是3 * 3的.. 转自:http://blog.csdn.net/spring371327/article/details/5297…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=3292 No more tricks, Mr Nanguo Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 587    Accepted Submission(s): 400 Problem Description Now Sailormoon…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4965 题意 给出两个矩阵 一个A: n * k 一个B: k * n C = A * B M = (A * B) ^ (n * n) 然后将M中所有的元素对6取余后求和 思路 矩阵结合律.. M = (A * B) * (A * B) * (A * B) * (A * B) * (A * B) * (A * B) * (A * B) * (A * B) -- 其实也等价于 M = A * (B *…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4565 题目大意: 给出a,b,n,m,求出的值, 解题思路: 因为题目中出现了开根号,和向上取整后求余,所以用矩阵快速幂加速求解过程的时候,会产生误差,就很自然地想到了凑数,因为(a-1)^2<b<a^2,得出0<a-sqrt(b)<1,则无论n取多大,(a-sqrt(b))^n都是小于1的,(a-sqrt(b))^n 与 (a+sqrt(b))^n共轭,两者展开后会相互抵销,所以(…