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[35][35];
};
Mat operator* (Mat x, Mat y)
{
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    for(int t = 1; t <= n; t++)
    {
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; 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 = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            c.mat[i][j] = (i == j);
    while(y)
    {
        if(y & 1)
            c = c * x;
        x = x * x;
        y >>= 1;
    }
    return c;
}
Mat operator + (Mat x, Mat y)
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; 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 == 1)
        return temp;
    int mid = t / 2;
    Mat c = dfs(mid, temp);
    if(t & 1)
    {
       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 = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            scanf("%d", &a.mat[i][j]);
    c = dfs(k,a);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j < n; j++)
            printf("%d ", c.mat[i][j]);
        printf("%d
", c.mat[i][n]);
    }
    return 0;
}