Matrix Power Series

最新推荐文章于 2020-10-11 21:31:57 发布

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本文介绍如何通过矩阵幂运算解决求矩阵幂次之和的问题，并应用快速幂算法优化计算效率，适用于矩阵操作和算法优化场景。

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 16743		Accepted: 7135

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

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

Source

POJ Monthly--2007.06.03, Huang, Jinsong

#include <cstdio>
#include <math.h>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;

struct matt
{
int m[75][75];
} ;

matt f,s,unitmatt;

int n,m;
int k;

matt mul(matt a,matt b) //矩阵相乘
{
int i,j,k;
matt c;

for (i=1;i<=2*n;i++)
{
for (j=1;j<=2*n;j++)
{
c.m[i][j] = 0;
for (k=1;k<=2*n;k++)
{
c.m[i][j]+=a.m[i][k]* b.m[k][j];
c.m[i][j]%=m;
}
}
}
return c;
}

matt pow_m(matt a,int kk) //矩阵快速幂
{

matt res = unitmatt;
while(kk)
{
if (kk&1)
{
res=mul(res,a);
}

a=mul(a,a);
kk=kk>>1;
}

return res;

}

int main()
{
int i,j;
scanf("%d%d%d",&n,&k,&m);
//////////////
for (i=1;i<=n*2;i++)
{
unitmatt.m[i][i]=1;
}

for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&f.m[i][j]);
}
}

for(i=1;i<=n;i++)
{
f.m[i][i+n]=1;
f.m[i+n][i+n]=1;
}
//////////////////////////////////////////////////构造B= |A E | 矩阵
|0 E |
s=pow_m(f,k+1);

for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(i!=j)
printf("%d ",s.m[i][j+n ]); //输出矩阵的右上子矩阵，注意减E。
else
{
if (s.m[i][j+n]==0)
printf("3 ");
else
printf("%d ",s.m[i][j+n]-1);
}
} printf("\n");
}

return 0;

}

这样构造矩阵的方法很巧妙，还可以用在例如求斐波那契数列等地方