[矩阵乘法] PKU3233 Matrix Power Series

[

矩

阵

乘

法

]

M

a

t

r

i

x

P

o

w

e

r

S

e

r

i

e

s

[矩阵乘法]Matrix Power Series

[矩阵乘法]MatrixPowerSeries

Description

Given a

n

×

n

n × n

n×n matrix

A

A

A and a positive integer

k

k

k, find the sum

S

=

A

+

A

2

+

A

3

+

.

.

.

+

A

k

S = A + A^2 + A^3 + ... + A^k

S=A+A2+A3+...+Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers

n

n

n (

n

n

n ≤

30

30

30),

k

k

k (

k

k

k ≤

1

0

9

10^9

109) and

m

m

m (

m

m

m <

1

0

4

10^4

104). Then follow n lines each containing

n

n

n nonnegative integers below

32

,

768

32,768

32,768, giving

A

A

A’s elements in row-major order.

Output

Output the elements of

S

S

S modulo m in the same way as

A

A

A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

---

题目解析

为了降低时间复杂度，考虑矩阵乘法

然后可以构造出一个

2

r

2r

2r阶的矩阵

T

T

T

∣

A

E

O

E

∣

\begin{vmatrix} A & E \\ O & E \\ \end{vmatrix}

∣∣∣∣​AO​EE​∣∣∣∣​

其中：

A

A

A为输入的矩阵（

A

A

A是

r

r

r阶的矩阵）

O

O

O为全零矩阵 （

O

O

O是值全为

0

0

0的

r

r

r阶矩阵）

E

E

E为对角线矩阵（

E

E

E是除了对角线为

1

1

1，其他的都为

0

0

0的矩阵）

然后可以得出：

∣

S

[

n

−

1

]

,

A

n

∣

=

∣

S

[

n

−

2

]

,

A

n

−

1

∣

∗

T

|S[n-1],A^n| = |S[n-2],A^{n-1}| * T

∣S[n−1],An∣=∣S[n−2],An−1∣∗T

然后通过将矩阵乘法的结合律通过快速幂来计算出

T

n

T^n

Tn再可

A

∗

T

n

A*T^n

A∗Tn来求得答案

---

关于

T

T

T矩阵的实现

//全零矩阵的实现
//matrix 是已经定义的结构体，n和m是表示矩阵的长和宽，t是矩阵的值
matrix O (int re)
{
	matrix c;
	c.n = c.m = re;
	for (int i = 1; i <= re; ++ i)
	 for (int j = 1; j <= re; ++ j)
	  c.t[i][j] = 0;
	return c;
}

//对角线矩阵的实现
//matrix 是已经定义的结构体，n和m是表示矩阵的长和宽，t是矩阵的值，O函数为前文定义的全零矩阵
matrix E (int re)
{
	matrix c;
	c.n = c.m = re;
	c = O (re);
	for (int i = 1; i <= re; ++ i)
	 c.t[i][i] = 1;
	return a;
}

//关于矩阵的合并。n，m，t，O(),E()前文已述，T1就是前文提到的T矩阵，re为前文提到的r，a是前文提到的A
matrix hb (int re)
{
	t1.n = t1.m = re * 2;
	for (int i = 1; i <= re; ++ i)
	 for (int j = 1; j <= re; ++ j)
	  t1.t[i][j] = a.t[i][j];
	matrix er = E (re);
	for (int i = 1; i <= re; ++ i)
	 for (int j = re + 1; j <= re * 2; ++ j)
	  t1.t[i][j] = er.t[i][j];
	for (int i = re + 1; i <= re * 2; ++ i)
	 for (int j = re + 1; j <= re * 2; ++ j)
	  t1.t[i][j] = er.t[i][j];
	for (int i = re + 1; i <= re * 2; ++ i)
	 for (int j = 1; j <= re; ++ j)
	  t1.t[i][j] = 0;
}