Uva10689 yet another number sequence (matrix fast power)

 

 

Ultraviolet A 10689 yet another number sequence (matrix fast power)

 

 

It's just the power of the matrix. There's nothing to say.

 

Analysis: it is actually a Fibonacci series. Only the final matrix corresponding to () is (a, B) mod = 1; for (INT I = 0; I <m; ++ I) mod * = 10;

 

Code

#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;}

Code Jun

 

Uva10689 yet another number sequence (matrix fast power)