CodeForces 893B Beautiful Divisors

最新推荐文章于 2021-01-25 18:03:51 发布

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#ACM #矩阵快速幂

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本文介绍了一种算法，用于解决给定一个 n×n 的矩阵 A 和一个正整数 k，如何找到矩阵幂级数 S = A + A^2 + A^3 + … + A^k 的和的问题。输入包括矩阵的大小 n、幂的最大值 k 以及用于取模的 m。文章提供了一个 C++ 实现示例，展示了如何通过快速幂运算来高效地计算矩阵的幂级数和。

C - Matrix Power Series
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`

#include<iostream>
#include<vector>
//#include<algorithm>
using namespace std;
typedef vector<int>vec;
typedef vector<vec>mat;
typedef long long ll;
int M;	
int n, K;
//计算A*B
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,ll n) {
	mat B(A.size(), vec(A.size()));
	for (int i = 0; i < A.size(); i++) {
		B[i][i] = 1;
	}
	while (n > 0) {
		if (n & 1)B = mul(B, A);
		A = mul(A, A);
		n >>= 1;
	}
	return B;
}

mat add(mat a,mat b) {
	mat c(a.size(), vec(b.size()));
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			c[i][j] = (a[i][j] + b[i][j])%M;
	return c;
}


int main() {
	int val;
	cin >> n >> K >> M;
	mat A(n),B(n),tmp(n);
	for (int i = 0; i < n; i++){
		for (int j = 0; j < n; j++) {
			cin >> val;
			A[i].push_back(val);
			B[i].push_back(0);
		}
	}
	for (int i = 1; i <= K; i++)
	{
		tmp = pow(A, i);
		B = add(tmp, B);
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			cout << B[i][j]<<" ";
		}
		cout << endl;
	}
	system("pause");
	return 0;
}