【矩阵快速幂】 Blocks POJ3746

3744:Blocks

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he start to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.

输入

The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.

输出

For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.

样例输入

2
1
2

样例输出

2
6

来源

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17), Simon

 

 

#include <iostream>
using namespace std;

const int mn = 1e9 + 10, mod = 10007;

/// 均为偶数 a[i + 1] = 2 * a[i] + b[i];
/// 其中一个为偶数 b[i + 1] = 2 * (a[i] + b[i] + c[i])
/// 均为偶数 c[i + 1] = b[i] + 2 * c[i]
/// 构造矩阵a
int a[3][3] = {{2, 1, 0}, {2, 2, 2}, {0, 1, 2}};

struct node  /// 结构体内保存数组 方便函数递归及计算
{
	int c[3][3];
};

node mul(const node& p, const node& q)
{
	node t;
	for (int i = 0; i < 3; i++)
		for (int j = 0; j < 3; j++)
		t.c[i][j] = 0;

	for (int i = 0; i < 3; i++)  // 矩阵相乘
		for (int j = 0; j < 3; j++)
		for (int k = 0; k < 3; k++)
		t.c[i][j] = (t.c[i][j] + p.c[i][k] * q.c[k][j]) % mod;
	return t;
}

node pow(int n)
{
	node p, q;  /// q = 满足 (q * p = p) 的矩阵
	q.c[0][0] = 1, q.c[0][1] = 0, q.c[0][2] = 0;
	q.c[1][0] = 1, q.c[1][1] = 0, q.c[1][2] = 1;
	q.c[2][0] = 0, q.c[2][1] = 0, q.c[2][2] = 1;
	for (int i = 0; i < 3; i++)
		for (int j = 0; j < 3; j++)
			p.c[i][j] = a[i][j];

	while (n > 0)  // 快速幂
	{
		if (n & 1)
			q = mul(q, p);
		p = mul(p, p);
		n >>= 1;
	}
	return q;
}

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int n;
		scanf("%d", &n);

        node q = pow(n - 1);  // 计算矩阵自身相乘(n - 1)次

        int d[3] = {0};  // 答案矩阵
        int b[3] = {2, 2, 0}; /// 初始矩阵 n = 1时
        for (int i = 0; i < 3; i++)
		{
			for (int j = 0; j < 3; j++)
				d[i] = (d[i] + q.c[i][j] * b[j]) % mod;
		}

		printf("%d\n", d[0]);
	}
	return 0;
}