POJ3233_Matrix Power Series_递推法|矩阵快速幂

Matrix Power Series

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 22749 Accepted: 9503

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 40 11 1

Sample Output

1 22 3

给定 n*n 的矩阵 A 和正整数 k 和 m。求矩阵 A的幂的和。

递推法结合矩阵快速幂求解。

#include<cstdio>#include<iostream>#include<vector>using namespace std;typedef vector<int> vec;typedef vector<vec> mat;int n, k, m;mat Mul(mat A, mat B){mat C(A.size(), vec(B[0].size()));for(int i= 0; i< A.size(); i++)for(int k= 0; k< B.size(); k++)for(int j= 0; j< B[0].size(); j++)C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % m;return C;}mat Pow(mat A, int n){mat C(A.size(), vec(A.size()));for(int i= 0; i< A.size(); i++)C[i][i] = 1;while(n){if(n & 1) C = Mul(C, A);A = Mul(A, A);n = n >> 1;}return C;}int main(){scanf("%d %d %d", &n, &k, &m);//构造输入矩阵mat A(n, vec(n));for(int i= 0; i< n; i++)for(int j= 0; j< n; j++)scanf("%d", &A[i][j]);//构造初始矩阵mat B(n*2, vec(n));for(int i= 0; i< n; i++)for(int j= 0; j< n; j++)B[i+n][j] = A[i][j];//构造幂乘矩阵mat C(n*2, vec(n*2));for(int i= 0; i< n; i++)C[i][i] = C[i][i+n] = 1;for(int i= 0; i< n; i++)for(int j= 0; j< n; j++)C[i+n][j+n] = A[i][j];//最终矩阵B = Mul(Pow(C, k), B);for(int i= 0; i< n; i++){printf("%d", B[i][0]);for(int j= 1; j< n; j++)printf(" %d", B[i][j]);printf("\n");}return 0;}