HDU3306

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 <= 2 31 – 1
X : 2<= X <= 2 31– 1
Y : 2<= Y <= 2 31 – 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

裸地矩阵快速幂
虽然是这样不过推还是用了很久。。

#include<iostream>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<cstring>#include<cstdio>using namespace std;int mo = 10007;struct p{    int q[4][4];};p cheng(p q, p w){    p fh;    for (int a = 0; a < 4; a++)    {        for (int b = 0; b < 4; b++)        {            fh.q[a][b] = 0;            for (int c = 0; c < 4; c++)            {                fh.q[a][b] += (q.q[a][c]%mo * w.q[c][b]%mo+mo ) % mo;                fh.q[a][b] = (fh.q[a][b]+mo) % mo;            }        }    }    return fh;}p ksm(p ds, int zs){    p fs;    for (int a = 0; a < 4; a++)    {        for (int b = 0; b < 4; b++)        {            if (a != b)fs.q[a][b] = 0;            else fs.q[a][b] = 1;        }    }    while (zs)    {        if (zs & 1)fs = cheng(fs, ds);        ds = cheng(ds, ds);        zs >>= 1;    }    return fs;}int  main(){    int n, x, y;    while (scanf("%d%d%d", &n, &x, &y) == 3)    {        x %= mo, y %= mo;        p ds = { {1,x*x%mo,y*y%mo,2*x*y%mo,0,x*x%mo,y*y%mo,2*x*y%mo,0,1,0,0,0,x,0,y} };        ds = ksm(ds, n - 1);        printf("%d\n",(2*ds.q[0][0]%mo + ds.q[0][1]%mo + ds.q[0][2]%mo + ds.q[0][3]%mo+4*mo ) % mo );    }}