【POJ3233】【矩阵乘法】【快速幂】Matrix Power Series

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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 follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

题目大意

对于任意给定的$r$阶方阵$A$，求$S[n]=A+A^2+A^3+\dots +A^n$

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解题思路

构造一个1*2的矩阵
$J_1=\left[\begin{array}{cc}{A}^{n-1}& S[n-2]\end{array}\right]$

构造一个2*2的矩阵$J_2$
使得$J_1=\left[\begin{array}{cc}{A}^{n-1}& S[n-2]\end{array}\right]\ast J_2=\left[\begin{array}{cc}{A}^n& S[n-1]\end{array}\right]$

那么$J_2=\left[\begin{array}{cc}A& E\\ O& E\end{array}\right]$
O：全零矩阵
E（单位矩阵）：主对角线上是1（任何一个矩阵乘上单位矩阵都等于它本身）（随便列两个矩阵都可以验证的）

那么矩阵套矩阵怎么算呢，肯定不是矩阵乘法套矩阵乘法这么弱智啊~~(我就是那个弱智)~~
把矩阵拆开来，左上角是A，右上角和右下角是单位矩阵，左下角是全零矩阵，矩阵大小$(n\ast 2)\ast (n\ast 2)$以
以$n=3$为例
$\left[\begin{array}{cccccc}{A}_{1,1}& A_{1,2}& A_{1,3}& 1& 0& 0\\ A_{2,1}& A_{2,2}& A_{2,3}& 0& 1& 0\\ A_{3,1}& A_{3,2}& A_{3,3}& 0& 0& 1\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]$

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Code

#include <iostream>
#include <cstring>
#include <cstdio>
#define ll long long

using namespace std;

struct DT{
    
    
	ll n, m;
	ll aed[70][70];
}A, B, Bc;
ll n, Mod, k;

DT operator * (DT a, DT b){
    
    
	DT c;
	memset (c.aed, 0, sizeof (c.aed));
	c.n = a.n, c.m = b.m;
	for (int k = 1; k <= a.m; k++)
		for (int i = 1; i <= c.n; i++)
			for (int j = 1; j <= c.m; j++)
				c.aed[i][j] = (c.aed[i][j] + a.aed[i][k] * b.aed[k][j] % Mod) % Mod;
	return c;
}

void power (ll x){
    
    
	if (x == 1)
	{
    
    
		B = Bc;
		return;
	}
	power (x / 2);
	B = B * B;
	if (x % 2)
		B = B * Bc;
}

int main(){
    
    
	scanf ("%lld%lld%lld", &n, &k, &Mod);
	A.n = A.m = Bc.m = Bc.n = n * 2;
	//A = A 和 J1；J2 = Bc和B
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
    
    
			scanf ("%lld", &A.aed[i][j]);
			Bc.aed[i][j] = A.aed[i][j];//J2的左上角是A数组
		}
	for (int i = 1; i <= n; i++)
		Bc.aed[i][i + n] = Bc.aed[i + n][i + n] = 1;//J2的右上角和右下角是单位矩阵
	power (k);
	A = A * B;
	for (int i = 1; i <= n; i++)
	{
    
    
		for (int j = n + 1; j <= n * 2; j++)
			printf ("%lld ", A.aed[i][j]);
		printf ("\n");
	}
}