Hdu 3326 Another kind of Fibonacci (matrix Construction)

Question:

Describes another Fibonacci

F [n] = x * F [n-1] + y * F [N-2];

Calculate segma (F [I] ^ 2 );

Train of Thought Analysis:

For details about the construction matrix, stamp me

The construction matrix can be obtained.

The intermediate matrix is

1 1 00

0 x ^ 2 y ^ 2 2 x * y

0 1 0 0

0x0 y

#include 
 
  #include 
  
   #include #include 
   
    #include 
    
     #define N 30typedef long long LL;using namespace std;const LL mod = 10007;struct matrix{    int a[4][4];}origin;matrix multiply(matrix x,matrix y){    matrix temp;    memset(temp.a,0,sizeof(temp.a));    for(int i=0;i<4;i++)    {        for(int j=0;j<4;j++)        {            for(int k=0;k<4;k++)            {                temp.a[i][j]+=x.a[i][k]*y.a[k][j];                temp.a[i][j]=(temp.a[i][j])%mod;            }        }    }    return temp;}matrix matmod(matrix a,LL k){    matrix res;    memset(res.a,0,sizeof res.a);    for(int i=0;i<4;i++)res.a[i][i]=1;    while(k)    {        if(k&1)        res=multiply(res,a);        k>>=1;        a=multiply(a,a);    }    return res;}int main(){    int  n,x,y;    while(scanf("%d%d%d",&n,&x,&y)!=EOF)    {        matrix st;        x%=mod;        y%=mod;        st.a[0][0]=st.a[0][1]=1;        st.a[0][2]=st.a[0][3]=0;        st.a[1][0]=0;        st.a[1][1]=(x*x)%mod;        st.a[1][2]=(y*y)%mod;        st.a[1][3]=(2*x*y)%mod;        st.a[2][0]=st.a[2][2]=st.a[2][3]=0;        st.a[2][1]=1;        st.a[3][0]=st.a[3][2]=0;        st.a[3][1]=x%mod;        st.a[3][3]=y%mod;        matrix end = matmod(st,n);        int ans=0;        for(int i=0;i<4;i++)        {            ans+=end.a[0][i];            ans%=mod;        }        printf("%d\n",ans%mod);    }    return 0;}