POJ 3233 Matrix Power Series (矩阵&快速等比数列求和取模)

最新推荐文章于 2021-02-26 01:31:21 发布

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最新推荐文章于 2021-02-26 01:31:21 发布

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分类专栏： POJ acm之路--数学文章标签：数论 POJ acm 矩阵

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Matrix Power Series

http://poj.org/problem?id=3233

Time Limit: 3000MS

Memory Limit: 131072K

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

题意要求矩阵S=A+A^2+A^3+...+A^k mod m，可以用二分的方法。

单个数的快速幂请参考这篇文章。
快速等比数列求和：如何求k=19的情况呢？首先根据19的二进制10011构造这么一阶差分序列：1,2,16，所以原序列是0,1,3,19.

那怎么求A+A^2+A^3+...+A^19呢？我们就按照原序列来求，也就要先求一阶差分序列。

由于第一个数是0，我们初始化ans=0

第二个数是1，ans=ans*A^1+A=A

第三个数是3，ans=ans*A^2+A(I+A)=A+A^2+A^3

由于二进制中间有两个0，所以与ans相乘的数变为A^16(算法：A^2*A^2=A^4，A^4*A^4=A^16)

与ans*A^16相加的数变为A+...A^16(算法：(A+A^2)*((I+A^2)=A+...+A^4,(A+...+A^4)*(I+A^4)=A+...+A^8,(A+...+A^8)*(I+A^8)=A+...A^16)

第四个数是19，ans=ans*A^16+(A+...A^16)

完整代码：

/*79ms,144KB*/

#include <cstdio>
#include <cstring>
const int matSize = 31;

int calSize = matSize, mod = 1000000007;

struct Matrix
{
	int val[matSize][matSize];

	Matrix(bool Init = false)///使用默认实参以减少代码量
	{
		for (int i = 0; i < calSize; i++)
		{
			for (int j = 0; j < calSize; j++)
				val[i][j] = 0;
			if (Init)
				val[i][i] = 1;///单位矩阵
		}
	}

	void print()
	{
		for (int i = 0; i < calSize; i++)
		{
			for (int j = 0; j < calSize; j++)
			{
				if (j)
					putchar(' ');
				printf("%d", val[i][j]);
			}
			puts("");
		}
	}
} Base;

Matrix operator + (Matrix &_a, Matrix &_b)
{
	Matrix ret;
	for (int i = 0; i < calSize; i++)
		for (int j = 0; j < calSize; j++)
			ret.val[i][j] = (_a.val[i][j] + _b.val[i][j]) % mod;
	return ret;
}

Matrix operator * (Matrix &_a, Matrix &_b)
{
	Matrix ret = Matrix();
	for (int i = 0; i < calSize; i++)
		for (int k = 0; k < calSize; k++)
			if (_a.val[i][k])///优化一下
				for (int j = 0; j < calSize; j++)
				{
					ret.val[i][j] += _a.val[i][k] * _b.val[k][j];
					ret.val[i][j] %= mod;
				}
	return ret;
}

void deal(int k)
{
	Matrix one = Matrix(true), ans = Matrix();
	Matrix ep = Base, tmp = Base, temp;
	while (k)
	{
		if (k & 1)
		{
			ans = ans * ep;
			ans = ans + tmp;
		}
		temp = one + ep;
		tmp = tmp * temp;
		ep = ep * ep;
		k >>= 1;
	}
	ans.print();
}

int main()
{
	int k;
	while (~scanf("%d%d%d", &calSize, &k, &mod))
	{
		for (int i = 0; i < calSize; i++)
			for (int j = 0; j < calSize; j++)
			{
				scanf("%d", &Base.val[i][j]);
				Base.val[i][j] %= mod;
			}
		deal(k);
	}
	return 0;
}