hdu3306：Another kind of Fibonacci

A(0)=A(1)=1，A(i)=X*A(i-1)+Y*A(i-2)，求S(n)=A(0)^2+A(1)^2+A(2)^2+A(3)^2+……+A(n)^2。

这个矩阵有点毒。。

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<math.h>
 //#include<iostream>
 using namespace std;

 #define LL long long
 LL n,x,y;
 typedef LL mat[][];
 mat a,f,ans;
 const int mod=;
 void copy(mat &a,mat b)
 {
     for (int i=;i<=;i++)
         for (int j=;j<=;j++)
             a[i][j]=b[i][j];
 }
 void mul(mat a,mat b,mat &ans)
 {
     mat t;memset(t,,sizeof(t));
     for (int i=;i<=;i++)
         for (int j=;j<=;j++)
             for (int k=;k<=;k++)
                 t[i][j]=(t[i][j]+a[i][k]*b[k][j]%mod)%mod;
     copy(ans,t);
 }
 void init(mat &a)
 {
     memset(a,,sizeof(a));
     for (int i=;i<=;i++) a[i][i]=;
 }
 void pow(mat a,LL b,mat &ans)
 {
     mat tmp,last;copy(tmp,a);init(last);
     while (b)
     {
         if (b&) mul(last,tmp,last);
         mul(tmp,tmp,tmp);
         b>>=;
     }
     copy(ans,last);
 }
 int main()
 {
     while (~scanf("%lld%lld%lld",&n,&x,&y))
     {
         memset(a,,sizeof(a));
         a[][]=a[][]=a[][]=;
         a[][]=x*x%mod;
         a[][]=y*y%mod;
         a[][]=*x*y%mod;
         a[][]=x;a[][]=y;
         f[][]=f[][]=f[][]=f[][]=;
         pow(a,n,a);
         mul(a,f,ans);
         printf("%lld\n",ans[][]);
     }
     return ;
 }