矩阵幂求和

题意：

　　给矩阵A，求Σi=(1,k) Aⁱ 

SOL:

　　虽然可以快速幂，然而还是会炸，那么我们就优化一下。。。

　　

　　若k是偶数，则S=(1+1^(k/2))(A^1+A^2+……+A^(k/2))

　　若k是奇数，则S=(1+1^(k/2))(A^1+A^2+……+A^(k/2))+A^k

 

　　像是快速幂套快速幂。。。。复杂度。。一个log还是两个log呢。。

Code：

　　

#include <iostream>
#include <cstdio>
using namespace std;

int n, m;
struct Matrix
{
	int x[31][31];
};

void matrixPrint(Matrix mat)	
{
	for (int i = 0; i < n; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			if (j > 0) printf(" ");
			printf("%d", mat.x[i][j]);
		}
		printf("\n");
	}
}

Matrix matrixMultiply(Matrix a, Matrix b)	
{
	Matrix ret;
	for (int i = 0; i < n; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			ret.x[i][j] = 0;
			for (int t = 0; t < n; ++t)
			{
				ret.x[i][j] += a.x[i][t] * b.x[t][j];
				if (ret.x[i][j] >= m) ret.x[i][j] %= m;
			}
		}
	}
	return ret;
}

Matrix matrixPow(Matrix mat, int p)	
{
	Matrix ret;
	Matrix tmp = mat;
	for (int i = 0; i < n; ++i)
	{
		ret.x[i][i] = 1;
		for (int j = i + 1; j < n; ++j)
		{
			ret.x[i][j] = 0;
			ret.x[j][i] = 0;
		}
	}
	while (p > 0)
	{
		if (p & 1) ret = matrixMultiply(ret, tmp);
		p >>= 1;
		tmp = matrixMultiply(tmp, tmp);
	}
	return ret;
}

Matrix matrixSummary(Matrix mat, int k)
{
	for (int i = 0; i < n; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			if (mat.x[i][j] >= m) mat.x[i][j] %= m;
		}
	}

	if (k == 1) return mat;
	Matrix M1 = matrixPow(mat, k / 2);
	for (int i = 0; i < n; ++i)
	{
		M1.x[i][i] += 1;
	}
	Matrix M2 = matrixSummary(mat, k / 2);
	Matrix ret = matrixMultiply(M1, M2);
	if (k & 1)
	{
		Matrix tmp = matrixPow(mat, k);
		for (int i = 0; i < n; ++i)
		{
			for (int j = 0; j < n; ++j)
			{
				ret.x[i][j] += tmp.x[i][j];
				if (ret.x[i][j] >= m) ret.x[i][j] %= m;
			}
		}
	}
	return ret;
}

int main()
{
	Matrix mat;
	int k;
	scanf("%d%d%d", &n, &k, &m);
	for (int i = 0; i < n; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			scanf("%d", &mat.x[i][j]);
		}
	}
	Matrix ret = matrixSummary(mat, k);
	matrixPrint(ret);
	return 0;
}

 貌似还有更牛逼的。。。到时候再学下好了。。。