Codeforces 450B - Jzzhu and Sequences ( 矩阵快速幂 )

题意：

给定f1和f2，求fn

分析:

特判f1,f2

当n>=3时使用矩阵快速幂即可( 简单题 )

将公式转化一下 , 可以得到一个变换矩阵

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR( a, b )    memset( a, b, sizeof(a) )
#define MAT_SIZE 2
#define MOD 1000000007
typedef long long LL;
struct Mat
{
    LL mat[MAT_SIZE][MAT_SIZE];
    void zero()
    {
        for( int i = 0; i < MAT_SIZE; ++i )
            for( int j = 0; j < MAT_SIZE; ++j )
                mat[i][j] = 0;
    }
    void init()
    {
        for( int i = 0; i < MAT_SIZE; ++i )
            for( int j = 0; j < MAT_SIZE; ++j )
                mat[i][j] = ( i == j );
    }
    void setv( int v )
    {
        for( int i = 0; i < MAT_SIZE; ++i )
            for( int j = 0; j < MAT_SIZE; ++j )
                mat[i][j] = v;
    }
    Mat operator*( const Mat &b)const
    {
        Mat c;
        c.zero();
        for( int i =0; i < MAT_SIZE; ++i )
            for( int j =0; j < MAT_SIZE; ++j )
                for( int k = 0; k < MAT_SIZE; ++k )
                    c.mat[i][j] = ( c.mat[i][j] + mat[i][k] * b.mat[k][j] ) % MOD;
        return c;
    }
};

Mat fast_mod( Mat a, int b )
{
    Mat res;
    res.init();
    while( b )
    {
        if( b & 1 ) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

void Orz()
{
    int n;
    LL x, y;
    while( ~scanf( "%lld %lld %d", &x, &y, &n ) )
    {
        if( n == 1 )    printf( "%lld\n", ( x % MOD + MOD ) % MOD );
        else if( n == 2 )    printf( "%lld\n", ( y % MOD + MOD ) % MOD );
        else
        {
            Mat t;
            t.mat[0][0] = 0;
            t.mat[0][1] = 1;
            t.mat[1][0] = -1;
            t.mat[1][1] = 1;
            t = fast_mod( t, n-2 );
            LL res = ( ( t.mat[1][0] * x + t.mat[1][1] * y ) % MOD + MOD ) % MOD;
            printf( "%lld\n", res );
        }
    }
}

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

代码君