矩阵快速幂(推导+模板+例题详解)

整数快速幂：

分解成二进制形式易得程序

int fastpow(int base,int n,int mod){int ans=1;while(n){if(n&1) ans*=base%mod;base*=base;n>>=1;}return ans%mod;
}

快速幂复杂度是O(logn)，不用快速幂是O(n)

矩阵快速幂：
把整数乘法改成矩阵乘法，原理一样

struct Mat{double m[maxn+5][maxn+5];Mat(){memset(m,0,sizeof m);//构造函数初始化为0阵}void build(){//建单位矩阵for(int i=0;i<=maxn;i++){m[i][i]=1;}}
};
Mat operator*(Mat &a,Mat &b){Mat t;for(int i=1;i<=N;i++){for(int j=1;j<=N;j++){for(int k=1;k<=N;k++){t.m[i][j]+=a.m[i][k]*b.m[k][j];}}}return t;
}void ksm(int n){//和整数快速幂格式完全一样ans.build();while(n>0){if(n&1){ans=ans*a;}a=a*a;n>>=1;}
}

模板题：
https://codeforces.com/gym/102966/problem/C
C. CLETS Patrols
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
The Corporation Of Ludicrously Evil Topography and Scouring (CLETS) is working on a new way to organize their henchmen’s patrols so that they will be impossible to predict for a certain handsome british spy that would try to infiltrate their base. To that end, when a henchman reaches a post, a computer will randomly pick to which post the henchman should go next.

By some classified means, we have acquired the probability distributions that the computers use to choose where to send the henchmen next. Your task is to use this information to determine, for every post, the probability that a guard will be there after M steps, so that our friend, the handsome british spy, can plan the safest route to the CLETS base.

A step is defined as moving from one post to another, or waiting in the same post, should the computer decide that. The Guard start their shift on the first post.

Input
The fist line contains two integers separated by a space, N and M where (1≤N≤200) and (1≤M≤10000): the number of posts, and the number of steps we are interested in. The next N lines contain exactly N numbers separated by a space where the j-th number on line i+1 represents the probability that the computer will send a guard to the j-th post from the i-th post.

Output
Print N lines; in the i-th line, print the probability that a guard will be in post number i after M steps.

Your answer is considered correct if its absolute or relative error doesn’t exceed 10−4.

Example
inputCopy
2 2
0.75 0.25
0.5 0.5
outputCopy
0.6875
0.3125

题意：给出一个有向完全图<V,E>，V中有N个顶点，从1号顶点出发跳转M次,每次从顶点vi跳转到顶点vj的概率为<vi,vj>，求最终落在各个顶点的概率。
思路：通过观察找到矩阵的递推关系（step=1，2，3…），通过递推关系找到通项公式，然后用快速幂解出。

#include <bits/stdc++.h>
const int maxn=250;
using namespace std;
int N,M;
struct Mat{double m[maxn+5][maxn+5];Mat(){memset(m,0,sizeof m);}void build(){for(int i=0;i<=maxn;i++){m[i][i]=1;}}
}a,a2,ans;
Mat operator*(Mat &a,Mat &b){Mat t;for(int i=1;i<=N;i++){for(int j=1;j<=N;j++){for(int k=1;k<=N;k++){t.m[i][j]+=a.m[i][k]*b.m[k][j];}}}return t;
}void ksm(int n){ans.build();while(n>0){if(n&1){ans=ans*a;}a=a*a;n>>=1;}
}int main()
{//cout << "Hello world!" << endl;cin>>N>>M;for(int i=1;i<=N;i++){for(int j=1;j<=N;j++){cin>>a.m[i][j];a2.m[i][j]=a.m[i][j];}}ksm(M-1);for(int i=1;i<=N;i++){double temp=0;for(int j=1;j<=N;j++){temp+=ans.m[1][j]*a2.m[j][i];}printf("%.4f\n",temp);}return 0;
}

矩阵快速幂(推导+模板+例题详解)相关推荐

【洛谷P3390】矩阵快速幂（模板）
贴一下矩阵快速幂的模板 #include<iostream> #include<cstdio> #include<cstring> #include<stri ...
快速幂+矩阵快速幂（总结+例题）
1.快速幂以求a的b次方来介绍: 首先把b转换成二进制数该二进制数第i位的权为 2^i - 1 . 比如 : 11的二进制是1011 11 = 2³×1 + 2²×0 + 2¹×1 + 2º×1 ...
【面试相关】python实现快速幂取余算法详解
假设我们要计算 2102^{10}210 对1000取模的结果,可以很简单的得到24.但是如果要求 210002^{1000}21000 对1000取模的结果,常规方法就行不通了,因为常规的变量无法容 ...
【网格图软判决译码】基于比特级的MAP译码(Bitwise MAP Decoding)算法原理推导和例题详解(intrinsic和extrinsic值)
Bitwise MAP Decoding算法详解 C C C表示一个(n,k)码,生成矩阵为 G G G,编码如下 Encoding u =
E. Product Oriented Recurrence （矩阵快速幂新模板）
E. Product Oriented Recurrence time limit per test 1 second memory limit per test 256 megabytes inpu ...
矩阵快速幂以及其优化【华东交大课程】
矩阵快速幂以及其优化[华东交大课程] 快速幂基础:C++快速幂_Kicamon的博客-CSDN博客矩阵快速幂就是在快速幂的基础上结合矩阵运算的用法,其用途较为广泛,可以很大程度上优化代码. 一.矩阵 ...
HDU 2256（矩阵快速幂）
传送门题面: Problem of Precision Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K ( ...
洛谷P3758/BZOJ4887 [TJOI2017] 可乐 [矩阵快速幂]
洛谷传送门,BZOJ传送门可乐 Time Limit: 10 Sec Memory Limit: 64 MB Submit: 299 Solved: 207 Description 加里敦星球的 ...
hdu 2842 Chinese Rings 矩阵快速幂
分析: 后面的环能不能取下来与前面的环有关,前面的环不被后面的环所影响.所以先取最后面的环设状态F(n)表示n个环全部取下来的最少步数先取第n个环,就得使1~n-2个环属于被取下来的状态,第n-1 ...
又见斐波那契~矩阵快速幂入门题
链接:https://www.nowcoder.com/acm/contest/105/G 来源:牛客网题目描述这是一个加强版的斐波那契数列. 给定递推式求F(n)的值,由于这个值可能太大,请对 ...

矩阵快速幂(推导+模板+例题详解)

矩阵快速幂(推导+模板+例题详解)相关推荐

最新文章

热门文章