[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}
∣∣∣∣AOEE∣∣∣∣
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;
}
