矩阵快速幂

矩阵快速幂

POJ3233

Matrix Power Series
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 23491 Accepted: 9773
Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.

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

题意：给一个 n * n 矩阵 ，计算 S = （A + A2 + A3 + … + Ak） % m

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int n , mol ;
struct martix{
    int m[50][50];
}ans ,base;


martix mult(martix x,martix y) 矩阵乘法
{
    martix tmp;
    for(int i = 0 ; i < n ; i++)
    for(int j = 0 ; j < n ; j++){
        tmp.m[i][j] = 0;
        for(int k = 0 ; k < n ; k++){
            tmp.m[i][j] = (tmp.m[i][j] + (x.m[i][k])*(y.m[k][j])) % mol;
        }
    }
    return tmp;
}


martix expo(martix x ,int k) 矩阵的幂
{
    martix tmp;
    int i , j;
    for(int i = 0 ; i < n ; i++)
    for(int j = 0 ; j < n ; j++){
        if(i == j) tmp.m[i][j] = 1;
        else tmp.m[i][j] = 0;
    }
    while(k){
        if(k & 1) tmp = mult(tmp , x);
        x = mult(x , x);
        k >>= 1;
    }
    return tmp;
}
martix add(martix a , martix b) 矩阵加法
{
    for(int i = 0 ; i < n ; i++)
    for(int j = 0 ; j < n ; j++)
        a.m[i][j] = (a.m[i][j] % mol + b.m[i][j] % mol ) % mol;
    return a;
}
martix sum( martix x, int k)
{
    martix tmp , y;
    if(k == 1) return x;
    tmp = sum(x , k / 2);
    if(k & 1){
        y = expo(x , k / 2 + 1);
        tmp = add(mult(y , tmp) , tmp);
        return add(tmp , y);
    }
    else{
        y = expo(x , k / 2);
        return add(mult(y , tmp) , tmp);
    }
}
int main()
{
    int k;
    while(~scanf("%d%d%d" , &n , &k , &mol)){
        for(int i = 0 ; i < n ; i++)
            for(int j = 0 ; j < n ; j++)
               scanf("%d",&base.m[i][j]);
        ans = sum(base ,k);
        for(int i = 0 ; i < n ; i++){
            for(int j = 0 ; j < n ; j++)
               printf("%d ",ans.m[i][j]);
            printf("\n");
        }
    }
    return 0;
}