【HDU】4965 Fast Matrix Calculation 矩阵快速幂

传送门：【HDU】4965 Fast Matrix Calculation

题目分析：因为比赛的时候写的太匆忙。。写的不堪入目，所以赛后重写了一次，顺便就贴一下了。

因为A*B=C，所以C^(N*N-1) = A*B*A*B*A*...*B*A*B，因为满足结合律所以变成A*( (B*A)^(N*N-2) )*B，因为中间得到的矩阵最大不超过K(K<=6)，所以可以对中间的矩阵快速幂，然后再暴力和左右两边的矩阵相乘即可。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )
#define FFF( a , b , c ) REP ( i , 0 , a ) REP ( j , 0 , b ) REP ( k , 0 , c )
#define FF( a , b ) REP ( i , 0 , a ) REP ( j , 0 , b )

const int MAXN = 1005 ;
const int K = 6 ;

int A[MAXN][K] , B[K][MAXN] , C[K][K] ;
int tmp1[MAXN][K] , tmp2[MAXN][MAXN] ;
int res[K][K] , mat[K][K] ;
int n , m ;

void solve () {
	int sum = 0 , nn = n * n - 1 ;
	CLR ( C , 0 ) ;
	CLR ( mat , 0 ) ;
	CLR ( tmp1 , 0 ) ;
	CLR ( tmp2 , 0 ) ;
	FF ( n , m ) scanf ( "%d" , &A[i][j] ) ;
	FF ( m , n ) scanf ( "%d" , &B[i][j] ) ;
	FFF ( m , m , n ) mat[i][j] = ( mat[i][j] + B[i][k] * A[k][j] ) % K ;
	REP ( i , 0 , m ) C[i][i] = 1 ;
	while ( nn ) {
		if ( nn & 1 ) {
			CLR ( res , 0 ) ;
			FFF ( m , m , m ) res[i][j] = ( res[i][j] + C[i][k] * mat[k][j] ) % K ;
			FF ( m , m ) C[i][j] = res[i][j] ;
		}
		CLR ( res , 0 ) ;
		FFF ( m , m , m ) res[i][j] = ( res[i][j] + mat[i][k] * mat[k][j] ) % K ;
		FF ( m , m ) mat[i][j] = res[i][j] ;
		nn >>= 1 ;
	}
	FFF ( n , m , m ) tmp1[i][j] = ( tmp1[i][j] + A[i][k] * C[k][j] ) % K ;
	FFF ( n , n , m ) tmp2[i][j] = ( tmp2[i][j] + tmp1[i][k] * B[k][j] ) % K ;
	FF ( n , n ) sum += tmp2[i][j] ;
	printf ( "%d\n" , sum ) ;
}

int main () {
	while ( ~scanf ( "%d%d" , &n , &m ) && ( n || m ) ) solve () ;
	return 0 ;
}