Matrix Power Series(POJ 3233)

• 原题如下：
  Matrix Power Series

  Time Limit: 3000MS		Memory Limit: 131072K
  Total Submissions: 28044		Accepted: 11440

  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

• 题解：构造矩阵：
  
  

  此时，令S_k=I+A+…+A^k-1，则有：
  
  通过计算这个矩阵的k次幂，就可求出A的累乘和，时间复杂度为O(n³logk)

• 代码：

   1 #include <cstdio>
   2 #include <cctype>
   3 #define number s-'0'
   4 #include <cstring>
   5 #include <vector>
   6 
   7 using namespace std;
   8 
   9 typedef vector<int> vec;
  10 typedef    vector<vec> mat;
  11 
  12 int n,k,m;
  13 mat A;
  14 
  15 void read(int &x)
  16 {
  17     char s;
  18     x=0;
  19     bool flag=0;
  20     while (!isdigit(s=getchar()))
  21         (s=='-')&&(flag=true);
  22     for (x=number; isdigit(s=getchar());x=x*10+number);
  23     (flag)&&(x=-x);
  24 }
  25 
  26 void write(int x)
  27 {
  28     if (x<0)
  29     {
  30         putchar('-');
  31         x=-x;
  32     }
  33     if (x>9) write(x/10);
  34     putchar(x%10+'0');
  35 }
  36 
  37 mat mul(mat &A, mat &B)
  38 {
  39     mat C(A.size(), vec(B[0].size()));
  40     for (int i=0; i<A.size(); i++)
  41     {
  42         for (int j=0; j<B[0].size(); j++)
  43         {
  44             for (int k=0; k<B.size(); k++)
  45             {
  46                 C[i][j]=(C[i][j]+A[i][k]*B[k][j])%m;
  47             }
  48         }
  49     }
  50     return C;
  51 }
  52 
  53 mat pow(mat A, int n)
  54 {
  55     mat B(A.size(), vec(A.size()));
  56     for (int i=0; i<A.size(); i++) B[i][i]=1;
  57     while (n>0)
  58     {
  59         if (n&1) B=mul(B, A);
  60         A=mul(A, A);
  61         n>>=1;
  62     }
  63     return B;
  64 }
  65 
  66 int main(int argc, char * argv[])
  67 {
  68     read(n);read(k);read(m);
  69     A=mat (n,vec(n));
  70     mat B(n*2, vec(n*2));
  71     for (int i=0; i<n; i++)
  72     {
  73         for (int j=0; j<n; j++)
  74         {
  75             read(A[i][j]);
  76             B[i][j]=A[i][j];
  77         }
  78         B[n+i][i]=B[n+i][n+i]=1;
  79     }
  80     B=pow(B,k+1);
  81     for (int i=0; i<n; i++)
  82     {
  83         for (int j=0; j<n; j++)
  84         {
  85             int a=B[n+i][j]%m;
  86             if (i==j) a=(a+m-1)%m;
  87             printf("%d%c", a, j+1==n?'\n':' ');
  88         }
  89     }
  90 }