poj3734(矩阵快速幂)

最新推荐文章于 2021-09-11 19:55:36 发布

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本文探讨了一个有趣的数学问题：如何计算在特定条件下（红色和绿色块数量均为偶数），一排N个方块可以用几种方式涂上红、蓝、绿、黄四种颜色。通过矩阵快速幂的方法，提供了一种高效的解决方案。

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7108		Accepted: 3440

Description

Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts 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.

Input

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.

Output

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.

Sample Input

2
1
2

Sample Output

2
6

Ai代表两个同时为偶数方案数，Bi表示两个中某一个为偶数方案数，Ci表示两个都为奇数方案数。

则有矩阵递推式：

Ai+1 2 1 0 Ai

Bi+1 = 2 2 2 Bi

Ci+1 0 1 2 Ci

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
//用二维vector来表示矩阵 
typedef vector<int> vec;
typedef vector<vec> mat;
typedef long long ll;

const int M=10007;       //模数 
//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;
}
//计算A^n 
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;
}
int n;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		mat A(3,vec(3));
		A[0][0]=2;
		A[0][1]=1;
		A[0][2]=0;
		A[1][0]=2;
		A[1][1]=2;
		A[1][2]=2;
		A[2][0]=0;
		A[2][1]=1;
		A[2][2]=2;
		mat B=pow(A,n);
		printf("%d\n",B[0][0]);
	}
}