hdu3306 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

#include <iostream>
#define mod 10007;
using namespace std;
const int Max=4;
typedef struct
{
long long m[Max][Max];
}Matrix;

Matrix P={1,1,0,0,
0,0,0,0,
0,1,0,0,
0,0,0,0};
Matrix I={1,0,0,0,
0,1,0,0,
0,0,1,0,
0,0,0,1};
Matrix matrixmul(Matrix a,Matrix b)
{
Matrix c;
for(int i=0;i<Max;i++)
for(int j=0;j<Max;j++)
{
c.m[i][j]=0;
for(int k=0;k<Max;k++)
{
c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;
}
c.m[i][j]=c.m[i][j]%mod;
}
return c;
}
Matrix quickpow(long long n)
{
Matrix m=P,b=I;
while(n>=1)
{
if(n&1)
b=matrixmul(b,m);
n=n>>1;
m=matrixmul(m,m);
}
return b;
}
int main()
{int n,x,y,sum;
Matrix tmp;
while(cin>>n>>x>>y)
{
sum=0;
x%=mod;
y%=mod;
P.m[1][1]=(x*x)%mod;
P.m[1][2]=(y*y)%mod;
P.m[1][3]=(2*x*y)%mod
P.m[3][1]=x;
P.m[3][3]=y;
tmp=quickpow(n);
for(int i=0;i<4;i++)
sum+=tmp.m[0][i]%mod;
long long ans=sum%mod;
cout<<ans<<endl;
}

return 0;
}