Another kind of Fibonacci

                                                       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 <= 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

主要是递推矩阵，然后矩阵快速幂：

根据S(n)=S(n-1)+A(n)^2，A(n)^2=X^2*A(n-1)^2+Y^2*A(n-2)^2+2*X*Y*A(n-1)*A(n-2)，A(n)*A(n-1)=Y*A(n-1)*A(n-2)+X*A(n-1)^2，可以构造如下的矩阵，然后用二分矩阵的方法求解即可。

递推矩阵

以上摘自：http://www.cnblogs.com/staginner/archive/2012/04/26/2471507.html

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod=10007;
struct matrix
{
    long long ma[5][5];
};
matrix multi(matrix x,matrix y)//矩阵相乘
{
    matrix ans;
    memset(ans.ma,0,sizeof(ans.ma));
    for(int i=1;i<=4;i++)
    {
        for(int j=1;j<=4;j++)
        {
            if(x.ma[i][j])//稀疏矩阵优化
            for(int k=1;k<=4;k++)
            {
                ans.ma[i][k]=(ans.ma[i][k]+(x.ma[i][j]*y.ma[j][k])%mod)%mod;
            }
        }
    }
    return ans;
}
matrix pow(matrix a,long long m)
{
    matrix ans;
    for(int i=1;i<=4;i++)
    {
        for(int j=1;j<=4;j++)
        {
            if(i==j)
            ans.ma[i][j]=1;
            else
            ans.ma[i][j]=0;
        }
    }
    while(m)
    {
        if(m&1)
        ans=multi(ans,a);
        a=multi(a,a);
        m=m>>1;
    }
    return ans;
}
int main()
{
    long long x,y,n;
    while(~scanf("%I64d%I64d%I64d",&n,&x,&y))
    {
        matrix a,b;
        memset(a.ma,0,sizeof(a.ma));
        memset(b.ma,0,sizeof(b.ma));
        a.ma[1][1]=1;
        a.ma[1][2]=1;
        a.ma[2][2]=(x*x)%mod;
        a.ma[2][3]=(y*y)%mod;
        a.ma[2][4]=(2*x*y)%mod;
        a.ma[3][2]=1;
        a.ma[4][2]=x;
        a.ma[4][4]=y;
        b.ma[1][1]=1;
        b.ma[2][1]=1;
        b.ma[3][1]=1;
        b.ma[4][1]=1;
        a=pow(a,n);
        a=multi(a,b);
        printf("%I64d\n",a.ma[1][1]);
    }
    return 0;
}