POJ3233 Matrix Power Series 矩阵乘法

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POJ3233 Matrix Power Series 矩阵乘法

http://poj.org/problem?id=3233 挺有意思的..学习到结构体作为变量的转移, 题意 : 给定矩阵A,求A + A^2 + A^3 + ... + A^k的结果(两个矩阵相加就是对应位置分别相加).输出的数据mod m.k<=10^9. http://blog.csdn.net/rowanhaoa/article/details/21024093 代码: #include<cstdio> #include<cstring> #include<i…

Poj 3233 Matrix Power Series(矩阵乘法)

Matrix Power Series Time Limit: 3000MS Memory Limit: 131072K Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + - + Ak. Input The input contains exactly one test case. The first line of input contains three po…

POJ3233 Matrix Power Series 矩阵快速幂矩阵中的矩阵

Matrix Power Series Time Limit: 3000MS Memory Limit: 131072K Total Submissions: 27277 Accepted: 11143 Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak. Input The input contains exactly one test ca…

POJ3233 [C - Matrix Power Series] 矩阵乘法

解题思路题目里要求\(\sum_{i=1}^kA^i\),我们不妨再加上一个单位矩阵,求\(\sum_{i=0}^kA^i\).然后我们发现这个式子可以写成这样的形式:\(A(A(A...)+E)+E)+E\)于是,我们可以将\(*A+E\)看做一次变换,然后尝试构造一个矩阵.我们发现: \[ (\left[ \begin{matrix} A & E \\ 0 & E \end{matrix} \right])^n= \left[ \begin{matrix} A^{n+1} &…

POJ3233:Matrix Power Series(矩阵快速幂+二分）

http://poj.org/problem?id=3233 题目大意:给定矩阵A,求A + A^2 + A^3 + … + A^k的结果(两个矩阵相加就是对应位置分别相加).输出的数据mod m.k<=10^9.这道题两次二分,相当经典.首先我们知道,A^i可以二分求出.然后我们需要对整个题目的数据规模k进行二分.比如,当k=6时,有:A + A^2 + A^3 + A^4 + A^5 + A^6 =(A + A^2 + A^3) + A^3*(A + A^2 + A^3)应用这个式子后,规模…

POJ3233 Matrix Power Series(矩阵快速幂+分治)

Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak. Input The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then…