Uva10689 Yet another Number Sequence ( 矩阵快速幂 )

 

 

Uva 10689 Yet another Number Sequence (  矩阵快速幂  )

 

 

题意：



就是矩阵快速幂，没什么好说的。

 

分析：



其实还是斐波那契数列。只是最后对应的矩阵不是(1,1)是(a,b)了

MOD = 1;

for( int i = 0; i < m; ++i )

    MOD *= 10;    

 

代码

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long LL;

#define MAX_SIZE 2

#define CLR( a, b ) memset( a, b, sizeof(a) )



LL a, b, n, m, t, MOD;



LL M[MAX_SIZE];

inline void scan( LL & x )

{

    char c;

    while( c = getchar(), c < '0' || c > '9' );

    x = c - '0';

    while( c = getchar(), c >= '0' && c <= '9' )    x = x*10 + c - '0';

}



struct Mat

{

    LL mat[MAX_SIZE][MAX_SIZE];



    void init()

    {

        for( int i = 0; i < 2; ++i )

            for( int j = 0; j < 2; ++j )

                mat[i][j] = ( i == j );

    }

    void zero()

    {

        for( int i = 0; i < 2; ++i )

            for( int j = 0; j < 2; ++j )

                mat[i][j] = 0;

    }

    void setv( int v )

    {

        for( int i = 0; i < 2; ++i )

            for( int j = 0; j < 2; ++j )

                mat[i][j] = v;

    }

    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;

        return c;

    }    

    void debug()

    {

        for( int i = 0; i < 2; ++i )

        {

            for( int j = 0; j < 2; ++j )

            {

                if( j != 0 )    putchar( ' ' );

                printf( "%lld", mat[i][j] );

            }

            putchar( '\n' );

        }

    }

};



Mat fast_mod( Mat a, LL b )

{

    Mat res;

    res.init();

    while( b )

    {

        if( b & 1 )    res = res * a;

        a = a * a;

        b >>= 1;

    }

    return res;

}



void Orz()

{

    Mat c;

    c.setv( 1 );

    c.mat[0][0] = 0;

    //c.debug();        

    scan( t );

    while( t-- )

    {

        scan( a ), scan( b ), scan( n ), scan( m );

        MOD = 1;

        for( int i = 0; i < m; ++i )

            MOD *= 10;    

        //M[0] = a, M[1] = b;

        if( n == 1)

            printf( "%lld\n", a % MOD );

        else if( n == 2 )

            printf( "%lld\n", b % MOD );

        else

        {    

            Mat res = fast_mod( c, n - 2 + 1 );

            LL ans = ( res.mat[1][0] * a + res.mat[1][1] * b ) % MOD;

            printf( "%lld\n", ans );

        }

    }

}



int main()

{

    Orz();    

    return 0;

}

代码君