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

Another kind of Fibonacci

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 <= 2³¹ – 1
X : 2<= X <= 2³¹ – 1
Y : 2<= Y <= 2³¹ – 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;
}