矩阵快速幂是基于普通的快速幂的一种扩展,如果不知道的快速幂的请参见http://www.cnblogs.com/Howe-Young/p/4097277.html.二进制这个东西太神奇了,好多优秀的算法都跟他有关系,这里所说的矩阵快速幂就是把原来普通快速幂的数换成了矩阵而已,只不过重载了一下运算符*就可以了,也就是矩阵的乘法,  当然也可以写成函数,标题中的这三个题都是关于矩阵快速幂的基础题.拿来练习练习熟悉矩阵快速幂,然后再做比较难点的,其实矩阵快速幂比较难的是构造矩阵.下面还是那题目直接说话…

Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 .Now, I will give a0 ~ a9 and two positiv…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4331    Accepted Submission(s): 2603 Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) =…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2791    Accepted Submission(s): 1659 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =…

一次ac 在做递推关系的题目的时候  快速幂矩阵真的很有用 #include<iostream> #include<cstdio> #include<cstring> using namespace std; int n,k; struct matrix{ ][]; }; matrix multi( matrix a, matrix b ) { matrix c; ;i<;i++) ;j<;j++){ c.arr[i][j]=; ;w<;w++) c.…

Lele now is thinking about a simple function f(x). If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 . Now, I will give a0 ~ a9 and two positive integers k and m…

对于数据量大的求余运算,在有递推式的情况下,可以构造矩阵求解. A - A Simple Math Problem Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1…

http://acm.hdu.edu.cn/showproblem.php?pid=1757 Problem Description Lele now is thinking about a simple function f(x).If x < 10  f(x) = x.If x >= 10  f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);And ai(0<=i<=9) can only be…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1757 按照题目的要求构造矩阵 //Author: xiaowuga //矩阵: //a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 9 // 1 0 0 0 0 0 0 0 0 0 8 // 0 1 0 0 0 0 0 0 0 0 7 // 0 0 1 0 0 0 0 0 0 0 6 // 0 0 0 1 0 0 0 0 0 0 5 // 0 0 0 0 1 0 0 0 0 0 4 //…

解题思路:分析需要不少时间,比较懒,直接把别人的分析贴在这里, 然后贴上自己写的代码: K相当之大.所以逐一递推的算法无法胜任.这时我们就不得不运用矩阵加速.首先来讲一下矩阵乘法: 若一矩阵的列数与另一矩阵的行数相等,则可定义这两个矩阵的乘积.如 A 是 m×n 矩阵和 B 是 n×p矩阵,它们是乘积 AB 是一个 m×p 矩阵,其中 (AB)[i, j] = A[i, 1] * B[1, j] + A[i, 2] * B[2, j] + ... + A[i, n] * B[n, j] 对所有…