HDU 2256 矩阵快速幂

题目链接

题意：

计算Floor((sqrt(2)+sqrt(3))^(2n))mod1024。

思路：

(sqrt(2)+sqrt(3))^2 = (5+sqrt(24))^2

因为(5+sqrt(24))和(5-sqrt(24))共轭，所以(5+sqrt(24))^n+(5-sqrt(24))^n为整数。

因为5-sqrt(24)<1，所以(5-sqrt(24))^n<1，所以Floor( (5+sqrt(24))^n ) = (5+sqrt(24))^n+(5-sqrt(24))^n-1。

定义Sn = (5+sqrt(24))^n+(5-sqrt(24))^n

Sn * ( (5+sqrt(24))+(5-sqrt(24)) ) 

= (5+sqrt(24))^(n+1)+(5+sqrt(24))^n*(5-sqrt(24))+(5-sqrt(24))^n*(5+sqrt(24))+(5-sqrt(24))^(n+1)

= (5+sqrt(24))^(n+1)+(5-sqrt(24))^(n+1)+(5*5-24)[(5+sqrt(24))^(n-1)+(5-sqrt(24))^(n-1)]

=S(n+1)+S(n-1)

==> S(n+1) = 10*Sn-S(n-1)

构造矩阵

Sn+1 [ 10 -1 ] Sn

Sn      [ 1    0  ] Sn-1

矩阵快速幂求解

C++代码：

#include<bits/stdc++.h>
using namespace std;
const int mod = 1024;

int a,b,n,m;

struct Matrix
{
    int M[2][2];
};

Matrix Multi( Matrix A , Matrix B )
{
    Matrix R;
    R.M[0][0] = 0; R.M[0][1] = 0;
    R.M[1][0] = 0; R.M[1][1] = 0;
    for ( int i=0 ; i<2 ; i++ )
        for ( int j=0 ; j<2 ; j++ )
            for ( int k=0 ; k<2 ; k++ )
                R.M[i][j] = ( R.M[i][j]+A.M[i][k]*B.M[k][j]%mod )%mod;
    return R;
}

Matrix Mqpow( Matrix E , int x )
{
    Matrix R;
    R.M[0][0] = 1; R.M[0][1] = 0;
    R.M[1][0] = 0; R.M[1][1] = 1;
    while ( x )
    {
        if ( x&1 ) R = Multi( R , E );
        x >>= 1; E = Multi( E , E );
    }
    return R;
}

int main()
{
    int T; scanf ( "%d" , &T );
    while ( T-- )
    {
        int n; scanf ( "%d" , &n );
        if ( n==1 )
            printf ( "9\n" );
        else if ( n==2 )
            printf ( "97\n" );
        else
        {
            Matrix E;
            E.M[0][0] = 10;
            E.M[0][1] = 1023;
            E.M[1][0] = 1;
            E.M[1][1] = 0;
            E = Mqpow( E , n-2 );
            printf ( "%d\n" , (E.M[0][0]*98%mod+E.M[0][1]*10%mod-1+mod)%mod );
        }
    }
    return 0;
}