POJ3233Matrix Power Series（十大矩阵问题之三 + 二分+矩阵快速幂）

http://poj.org/problem?id=3233

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 18658		Accepted: 7895

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

当n为奇数；假设 n = 7: A + A ^ 2 + A ^ 3 + A ^ 4 + A ^ 5 + A ^ 6 + A＾７ = A + A ^ 2 + A ^ 3 + A ^ 3 * (A + A ^ 2 + A ^ 3) + A ^ 7

当n为偶数的时候就简单了

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>

 using namespace std;
 int n,m,k;
 struct Mat
 {
     int mat[][];
 };
 Mat operator* (Mat x, Mat y)
 {
     Mat c;
     memset(c.mat, , sizeof(c.mat));
     for(int t = ; t <= n; t++)
     {
         for(int i = ; i <= n; i++)
         {
             for(int j = ; j <= n; j++)
                 c.mat[i][j] = (c.mat[i][j] + x.mat[i][t] % m * (y.mat[t][j] % m) ) % m;
         }
     }
     return c;
 }
 Mat operator^ (Mat x, int y)
 {
     Mat c;
     for(int i = ; i <= n; i++)
         for(int j = ; j <= n; j++)
             c.mat[i][j] = (i == j);
     while(y)
     {
         if(y & )
             c = c * x;
         x = x * x;
         y >>= ;
     }
     return c;
 }
 Mat operator + (Mat x, Mat y)
 {
     for(int i = ; i <= n; i++)
     {
         for(int j = ; j <= n; j++)
             x.mat[i][j] = ( x.mat[i][j] % m + y.mat[i][j] % m ) % m;
     }
     return x;
 }
 Mat dfs(int t,Mat temp)
 {
     if(t == )
         return temp;
     int mid = t / ;
     Mat c = dfs(mid, temp);
     if(t & )
     {
        c = c + (temp ^ mid ) * c;
        return c + (temp ^ t); //第一次交没加括号，查了好长时间的错，惭愧惭愧，其实codeblock都waring了，弱
     }
     else
         return c + (temp ^ mid) * c;
 }
 int main()
 {

     scanf("%d%d%d", &n,&k,&m);
     Mat a,c;
     for(int i = ; i <= n; i++)
         for(int j = ; j <= n; j++)
             scanf("%d", &a.mat[i][j]);
     c = dfs(k,a);
     for(int i = ; i <= n; i++)
     {
         for(int j = ; j < n; j++)
             printf("%d ", c.mat[i][j]);
         printf("%d\n", c.mat[i][n]);
     }
     return ;
 }