Modular Fibonacci

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=30506#problem/U

fib数列对2^m取模的循环节为3*（2^(m-1)）

 

#include<map>

#include<set>

#include<list>

#include<cmath>

#include<ctime>

#include<deque>

#include<stack>

#include<bitset>

#include<cstdio>

#include<vector>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<numeric>

#include<sstream>

#include<utility>

#include<iostream>

#include<algorithm>

#include<functional>



using namespace std ;

const int maxn = 1000005 ;



long long f[ 25 ][ maxn ] ;



void Union()

{

	for( int i = 1 ; i <= 20 ; ++i )

	{

		f[ i ][ 0 ] = 0 ;

		f[ i ][ 1 ] = 1 ;

		long long temp = 1 << i ;

		for( long long j = 2 ; j < 3 * ( 1 << ( i - 1 ) ) ; ++j )

		{

			f[ i ][ j ] = ( f[ i ][ j - 1 ] % temp + f[ i ][ j - 2 ] % temp ) % temp ;

		}

	}

} 



int main()

{

	Union() ;

	int m , n ;

	while( scanf( "%d%d" , &n , &m ) != EOF )

	{

		if( !m )

			printf( "0\n" ) ;

		else

			printf( "%lld\n" , f[ m ][ n % ( 3 * ( 1 << ( m - 1 ) ) ) ] ) ;

	}

    return 0;

}