poj 3233 Matrix Power Series 矩阵乘法

题意：对于任意给定的r阶方阵A，求S[n]=A+A²+A³+…+Aⁿ

^分析：

考虑1×2的矩阵【A^n-1,S[n-2]】，注意此1×2矩阵的2个元素都是r阶方阵！我们希望通过乘以某2×2的矩阵M，得到1×2的矩阵【Aⁿ,S[n-1]】＝【A^n-1*A, A^n-1+S[n-2]】

容易构造出此矩阵M为：

A	E
O	E

其中4个元素均为r阶方阵，O表示r阶全0矩阵，E表示单位矩阵（主对角线上为1，其它全为0）。然后用矩阵乘法快速幂算一下就好啦。

代码：

type
  arrr=array[1..30,1..30] of longint;
  arr=array[1..2,1..2,1..30,1..30] of longint;

var
  n,m,p,i,j,k:longint;
  ans,a,u:arrr;
  c,d:arr;

procedure cheng1(x,y:longint;a,b:arrr);
var
  i,j,k:longint;
begin
  for i:=1 to n do
    for j:=1 to n do
      for k:=1 to n do
        c[x,y,i,j]:=(c[x,y,i,j]+a[i,k]*b[k,j]) mod p;
end;

procedure cheng(a,b:arr);
var
  i,j,k:longint;
begin
  for i:=1 to 2 do
    for j:=1 to 2 do
    begin
      fillchar(c[i,j],sizeof(c[i,j]),0);
      for k:=1 to 2 do
        cheng1(i,j,a[i,k],b[k,j])
    end;
end;

procedure ksm(x:longint);
begin
  if x<=1 then exit;
  ksm(x div 2);
  cheng(c,c);
  if x mod 2=1 then cheng(c,d);
end;

begin
  readln(n,m,p);
  for i:=1 to n do
    for j:=1 to n do
      read(a[i,j]);
  for i:=1 to n do
    u[i,i]:=1;
  d[1,1]:=a; d[1,2]:=u; d[2,2]:=u;
  c:=d;
  ksm(m);
  for i:=1 to n do
    for j:=1 to n do
      for k:=1 to n do
        ans[i,j]:=(ans[i,j]+a[i,k]*c[1,2,k,j]) mod p;
  for i:=1 to n do
    for j:=1 to n do
      if j<n
        then write(ans[i,j],' ')
        else writeln(ans[i,j]);
end.