NEFU 458 矩阵连乘

http://acm.nefu.edu.cn/JudgeOnline/problemshow.php?problem_id=458

description

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

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
typedef long long LL;
const int N=4;
const int MOD=10007;
struct Matrix
{
    LL m[N][N];
};
Matrix I=
{1,0,0,0,
 0,1,0,0,
 0,0,1,0,
 0,0,0,1};
Matrix multi(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++)
        {
            c.m[i][j]=0;
            for(int k=0;k<N;k++)
            {
                c.m[i][j]+=a.m[i][k]*b.m[k][j]%MOD;
            }
            c.m[i][j]%=MOD;
        }
    return c;
}
Matrix quick_mod(Matrix a,LL k)
{
    Matrix ans=I;
    while(k!=0)
    {
        if(k&1)
        {
            ans=multi(ans,a);
        }
        k>>=1;
        a=multi(a,a);
    }
    return ans;
}
int main()
{
    LL x,y,z;
    while(~scanf("%lld%lld%lld",&z,&x,&y))
    {
        Matrix A={1,(x%MOD*x%MOD)%MOD,(y%MOD*y%MOD)%MOD,(2*x%MOD*y%MOD)%MOD,
                  0,(x%MOD*x%MOD)%MOD,(y%MOD*y%MOD)%MOD,(2*x%MOD*y%MOD)%MOD,
                  0,1,0,0,
                  0,x%MOD,0,y%MOD};
        Matrix ans=quick_mod(A,z-1);
        LL t=(ans.m[0][0]%MOD*2+ans.m[0][1]%MOD+ans.m[0][2]%MOD+ans.m[0][3]%MOD)%MOD;
        printf("%lld\n",t);
    }
    return 0;
}