HDU 4549 M斐波那契数列 ( 矩阵快速幂 + 费马小定理 )

HDU 4549 M斐波那契数列 (  矩阵快速幂 + 费马小定理  )

题意：中文题，不解释

分析：最好的分析就是先推一推前几项，看看有什么规律

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef __int64 LL;
#define CLR( a, b )    memset( a, b, sizeof(a) )
#define MOD 1000000007

struct Mat
{
    LL mat[2][2];
    Mat()
    {
        CLR( mat, 0 );
    }
    void init()
    {
        for( int i = 0; i < 2; ++i )
            for( int j = 0; j < 2; ++j )
                mat[i][j] = ( i == j );
    }
    void zero()
    {
        CLR( mat, 0 );
    }
    Mat operator * ( const Mat &b ) const
    {
        Mat c;
        c.zero();
        for( int k = 0; k < 2; ++k )
            for( int i = 0; i < 2; ++i )    if( mat[i][k] )
                for( int j = 0; j < 2; ++j )
                    c.mat[i][j] = ( c.mat[i][j] + mat[i][k] * b.mat[k][j] ) % ( MOD - 1 );
        return c;
    }
};

Mat Mat_mod( Mat a, LL b )
{
    Mat c;
    c.init();
    while( b )
    {
        if( b & 1 )    c = c * a;
        a = a * a;
        b >>= 1;
    }
    return c;
}

LL fast_mod( LL a, LL b )
{
    LL ans = 1;
    while( b )
    {
        if( b & 1 )    ans = ans * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ans;
}

void Orz()
{
    Mat c;
    c.mat[0][1] = c.mat[1][0] = c.mat[1][1] = 1;
    LL a, b, n;
    while( ~scanf( "%I64d %I64d %I64d",&a, &b, &n ) )
    {
        if( n == 0 )    printf( "%I64d\n", a % MOD );
        else if( n == 1 )    printf( "%I64d\n", b % MOD );
        else if( n == 2 )    printf( "%I64d\n", a * b % MOD );
        else
        {
            //0 1 1 2 3 5 8 13
            Mat t = Mat_mod( c, n - 3 );
            LL n1 = ( t.mat[1][0] + t.mat[1][1] ) % ( MOD - 1 );
            t = t * c;
            LL n2 = ( t.mat[1][0] + t.mat[1][1] ) % ( MOD - 1 );
            LL ans = fast_mod( a, n1 ) * fast_mod( b, n2 ) % MOD;
            printf( "%I64d\n", ans );
        }
    }
}

int main()
{
    Orz();
    return 0;
}

代码君