杭电oj3306：Another kind of Fibonacci（矩阵快速幂）

Another kind of Fibonacci

题目链接

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Problem Description

As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)^2 +A(1)^{2+……+A(n)}2.

Input

There are several test cases.

Each test case will contain three integers , N, X , Y .

N : 2<= N <= 2^31 – 1

X : 2<= X <= 2^31 – 1

Y : 2<= Y <= 2^31 – 1

Output

For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.

Sample Input

2 1 1

3 2 3

Sample Output

6

196

思路见下图:

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

const int maxn = 5;
typedef long long LL;

struct Matrix{
	int matrix[maxn][maxn];
}ori, ans;

int n = 4, N, X, Y, m = 10007;

void init()
{
	for(int i=0;i<n;i++)
		for(int j=0;j<n;j++)
			ori.matrix[i][j] = ans.matrix[i][j] = 0;
	ori.matrix[0][0] = ori.matrix[2][1] = 1;
	ori.matrix[0][1] = ori.matrix[1][1] = X * X % m;
	ori.matrix[0][2] = ori.matrix[1][2] = Y * Y % m;
	ori.matrix[0][3] = ori.matrix[1][3] = ((2 * X) % m) * Y % m;
	ori.matrix[3][1] = X;
	ori.matrix[3][3] = Y;
	ans.matrix[0][0] = 2;
	ans.matrix[1][0] = ans.matrix[2][0] = ans.matrix[3][0] = 1;
}

Matrix multiply(Matrix a, Matrix b)
{
	Matrix temp;
	memset(temp.matrix, 0, sizeof(temp.matrix));
	for(int i=0;i<n;i++)
		for(int j=0;j<n;j++)
			for(int k=0;k<n;k++)
				temp.matrix[i][j] = (temp.matrix[i][j] + ((a.matrix[i][k] * b.matrix[k][j]) % m)) % m;
	return temp;
}

//矩阵的b次幂
Matrix binaryPow(int b)
{
	Matrix temp;
	memset(temp.matrix, 0, sizeof(temp.matrix));
	for(int i=0;i<n;i++)
		temp.matrix[i][i] = 1;
	while(b > 0)
	{
		if(b & 1)
			temp = multiply(ori, temp);
		ori = multiply(ori, ori);
		b >>= 1;
	}
	return temp;
}

int main()
{
	//freopen("in.txt","r", stdin);
	while(cin>>N>>X>>Y)
	{
		X %= m;
		Y %= m;
		init();
		Matrix temp = binaryPow(N - 1);
		ans = multiply(temp, ans);
		cout<<ans.matrix[0][0]<<endl;
	}
	return 0;
}