讲的非常明白,看完后自己想了个矩阵就做了
a[n]=x*a[n-1]+y*a[n-2];
【a[n-1]^2, a[n-2]^2, a[n-1]*a[n-2], s[n-2]】*A矩阵=
【a[n]^2 , a[n-1]^2, a[n]*a[n-1], s[n-1]】=
【x^2*a[n-1]^2+y^2*a[n-2]+2*x*y*a[n-1]*a[n-2], a[n-1]^2, x*a[n-1]^2+y*a[n-1]*a[n-2], s[n-2]+a[n-1]^2】
矩阵A:
x^2 1 x 1
y^2 0 0 0
2*x*y 0 y 0
0 0 0 1
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
#define inf 0x3f3f3f3f
#define N 400010
#define ll __int64
struct node
{
ll a[4][4];
}I,T;
ll mod=10007;
node cheng(node A,node B)
{
node C;
memset(C.a,0,sizeof(C.a));
int i,j,k;
for(i=0;i<4;i++)
for(j=0;j<4;j++){
for(k=0;k<4;k++)
C.a[i][j]+=A.a[i][k]*B.a[k][j];
C.a[i][j]%=mod;
}
return C;
}
int main()
{
ll n,x,y;
while(cin>>n>>x>>y)
{
int i,j,k,l;
memset(T.a,0,sizeof(T.a));
memset(I.a,0,sizeof(I.a));
I.a[0][0]=I.a[1][1]=I.a[2][2]=I.a[3][3]=1;
T.a[0][0]=(x*x)%mod,T.a[1][0]=(y*y)%mod,T.a[2][0]=2*((x*y)%mod);
T.a[0][1]=T.a[0][3]=T.a[3][3]=1;
T.a[0][2]=x%mod,T.a[2][2]=y%mod;
while(n>0)
{
if(n&1) I=cheng(I,T);
n=n>>1;
T=cheng(T,T);
}
ll ans=(I.a[0][3]+I.a[1][3]+I.a[2][3]+I.a[3][3])%mod;
cout<<ans<<endl;
}
}