【POJ3233】Matrix Power Series

最新推荐文章于 2020-12-19 11:24:01 发布

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#矩阵乘法 #快速幂

博客围绕给定 n × n 矩阵 A 和正整数 k，求 S = A + A2 + A3 + … + Ak 展开。介绍了输入包含 n、k、m 及矩阵 A 元素，输出 S 各元素模 m 的结果。给出了从 ∣∣Ai−1Si−2∣∣ 转移到 ∣∣AiSi−1∣∣ 的思路及转移矩阵。

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

思路：

我们要将 $∣Ai−1Si−2∣\begin{vmatrix}A^{i-1}&S_{i-2}\end{vmatrix}$ 转移到 $∣AiSi−1∣\begin{vmatrix}A^{i}&S_{i-1}\end{vmatrix}$
那么就是 $∣Ai−1+ASi−2+Ai−1∣\begin{vmatrix}A^{i-1}+A&S_{i-2}+A^{i-1}\end{vmatrix}$
所以转移矩阵就是 $∣A101∣\begin{vmatrix}A&1\\0&1\end{vmatrix}$

code:

#include<iostream>
#include<cstdio>
using namespace std;
long long n, m, p;
long long a[900][900];
long long ans[900][900];
long long b[900][900];
void multi()
{
	long long c[100][100];
    for(long long i=1; i<=n*2; i++)
    	for(long long j=1; j<=n*2; j++)
    		c[i][j]=0;
    for(long long i=1; i<=n*2; i++)
    	for(long long j=1; j<=n*2; j++)
    		for(long long k=1; k<=n*2; k++)
    			c[i][j]=(c[i][j]+a[i][k]*ans[k][j])%p;
    for(long long i=1; i<=n*2; i++)
    	for(long long j=1; j<=n*2; j++)
    		ans[i][j]=c[i][j];
}
void multi1()
{
	long long c[100][100];
    for(long long i=1; i<=n*2; i++)
    	for(long long j=1; j<=n*2; j++)
    		c[i][j]=0;
    for(long long i=1; i<=n*2; i++)
    	for(long long j=1; j<=n*2; j++)
    		for(long long k=1; k<=n*2; k++)
    			c[i][j]=(c[i][j]+a[i][k]*a[k][j])%p;
    for(long long i=1; i<=n*2; i++)
    	for(long long j=1; j<=n*2; j++)
    		a[i][j]=c[i][j];
}
void ksm(long long k)
{
	for(long long i=1; i<=n*2; i++)
		ans[i][i]=1;
	while(k!=0)
	{
		if(k&1)
			multi();
		multi1();
		k>>=1; 
	}
}
void multi2()
{
	long long c[100][100];
    for(long long i=1; i<=n*2; i++)
    	for(long long j=1; j<=n*2; j++)
    		c[i][j]=0;
    for(long long i=1; i<=n; i++)
    	for(long long j=1; j<=n*2; j++)
    		for(long long k=1; k<=n*2; k++)
    			c[i][j]=(c[i][j]+b[i][k]*ans[k][j])%p;
    for(long long i=1; i<=n; i++)
    	for(long long j=1; j<=n*2; j++)
    		b[i][j]=c[i][j];
}
int main()
{
	scanf("%lld%lld%lld", &n, &m, &p);
	for(long long i=1; i<=n; i++)
		for(long long j=1; j<=n; j++)
			scanf("%lld", &a[i][j]), b[i][j]=a[i][j];
	for(long long i=1; i<=n; i++)
		for(long long j=n+1; j<=n*2; j++)
			if(i+n==j)
				a[i][j]=1;
	for(long long i=n+1; i<=n*2; i++)
		for(long long j=n+1; j<=n*2; j++)
			if(i==j)
				a[i][j]=1;
	ksm(m);
	multi2();
	for(long long i=1; i<=n; i++, cout<<endl)
		for(long long j=n+1; j<=n*2; j++)
			printf("%lld ", b[i][j]);
	return 0;
}