Another kind of Fibonacci

http://acm.hdu.edu.cn/showproblem.php?pid=3306

题解：

考虑1*4 的矩阵【s[n-2],a[n-1]^2,a[n-2]^2,a[n-1]*a[n-2]】

我们需要找到一个4×4的矩阵A，使得它乘以A得到1×4的矩阵

【s[n-1],a[n]^2,a[n-1]^2,a[n]*a[n-1]】

即：【s[n-2],a[n-1]^2,a[n-2]^2,a[n-1]*a[n-2]】* A = 【s[n-1],a[n]^2,a[n-1]^2,a[n]*a[n-1]】

= 【s[n-2]+a[n-1]^2 , x^2 * a[n-1]^2 + y^2 * a[n-2]^2 + 2*x*y*a[n-1]*a[n-2] ,a[n-1]^2 , x*a[n-1]^2 + y*a[n-2]a[n-1]】

可以构造矩阵A为：

1     0    0    0

1    x^2   1    x

0    y^2   0    0

0    2xy   0    y

故：【S[0],a[1]^2,a[0]^2,a[1]*a[0]】 * A^(n-1) = 【s[n-1],a[n]^2,a[n-1]^2,a[n]*a[n-1]】

所以：【S[0],a[1]^2,a[0]^2,a[1]*a[0]】 * A^(n) = 【s[n],a[n+1]^2,a[n]^2,a[n+1]*a[n]】

若A = (B * C ) 则AT = ( B * C )T = CT * BT

故

参考文章：矩阵乘积矩阵快速幂矩阵构造方法

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=10+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v,x,y;
int ans,cnt,flag,temp,sum;
int s[N][N];
int b[N][N];
int a[N][N];
char str;
struct node{};
void Matrix(ll a[N][N],ll b[N][N]){
    ll c[N][N];
    memset(c,0,sizeof(c));
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++){
            for(int k=0;k<4;k++){
                c[i][j]=(c[i][j]+a[i][k]*b[k][j])%10007;
            }
        }
    }
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++){
            a[i][j]=c[i][j];
        }
    }
}
ll power(int A,int B,int k){
    ll b[N][N]={{1,0,0,0},
                {1,A*A,1,A},
                {0,B*B,0,0},
                {0,2*A*B,0,B}};
    ll a[N][N]={{1,1,1,1},
                {0,0,0,0},
                {0,0,0,0},
                {0,0,0,0}};
    while(k){
        if(k&1)Matrix(a,b);
        Matrix(b,b);
        k>>=1;
    }
    return a[0][0];
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    while(~scanf("%d%d%d",&n,&x,&y)){
        cout<<power(x%10007,y%10007,n)<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}