BZOJ-1005: [HNOI2008]明明的烦恼（树的purfer编码）

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1005

树有种神奇的东西叫做purfer编码，每个编码对应y一棵树，然后就可以排列组合啦~

代码：

#include 
#include 
#include 
 
using namespace std ;
 
#define MAXN 10010
 
int p[ MAXN ] ;
 
void Init(  ) {
            bool f[ MAXN ] ; 
            memset( f , true , sizeof( f ) ) ;
            f[ 0 ] = f[ 1 ] = false , p[ 0 ] = 0 ;
            for ( int i = 1 ; i ++ < MAXN - 1 ; ) if ( f[ i ] ) {
                        p[ ++ p[ 0 ] ] = i ;
                        for ( int j = i << 1 ; j < MAXN ; j += i ) f[ j ] = false ;
            }
}
 
int n , sum = 0 , d[ MAXN ] , cnt = 0 ;
 
int num[ MAXN ] ;
 
void INC( int x ) {
            for ( int i = 0 ; i ++ < p[ 0 ] ; ) {
                        if ( x == 1 ) break ;
                        while ( ! ( x % p[ i ] ) ) ++ num[ i ] , x /= p[ i ] ;
            }
}
 
void DEC( int x ) {
            for ( int i = 0 ; i ++ < p[ 0 ] ; ) {
                        if ( x == 1 ) break ;
                        while ( ! ( x % p[ i ] ) ) -- num[ i ] , x /= p[ i ] ;
            }
}
 
struct bigint {
            
            int a[ MAXN ] , m ;
            
            bigint(  ) {
                        memset( a , 0 , sizeof( a ) ) ;
                        a[ 1 ] = m = 1 ;
            }
            
            void print(  ) {
                        printf( "%d" , a[ m ] ) ;
                        for ( int i = m - 1 ; i ; -- i ) {
                                     if ( a[ i ] < 1000 ) printf( "0" ) ;
                                     if ( a[ i ] < 100 ) printf( "0" ) ;
                                     if ( a[ i ] < 10 ) printf( "0" ) ;
                                     printf( "%d" , a[ i ] ) ;
                        }
                        printf( "\n" ) ;
            }
            
            void multi( int x ) {
                        for ( int i = 0 ; i ++ < m ; ) {
                                     a[ i ] *= x ;
                        }
                        int M = m ;
                        for ( int i = 0 ; i ++ < m ; ) {
                                     for ( int j = i ; a[ j ] >= 10000 ; ++ j ) {
                                                 a[ j + 1 ] += a[ j ] / 10000 ;
                                                 a[ j ] %= 10000 ;
                                                 M = max( M , j + 1 ) ;
                                     }
                        }
                        m = M ;
            }
            
} ans ;
 
int main(  ) {
            Init(  ) ;
            scanf( "%d" , &n ) ;
            for ( int i = 0 ; i ++ < n ; ) {
                        int x ; scanf( "%d" , &x ) ;
                        if ( n == 1 ) {
                                     if ( ! x ) printf( "1\n" ) ; else printf( "0\n" ) ;
                                     return 0 ;
                        }
                        if ( x != - 1 ) sum += ( x - 1 ) , d[ ++ cnt ] = x - 1 ;
            }
            if ( sum > n - 2 ) {
                        printf( "0\n" ) ; return 0 ;
            }
            if ( n == cnt ) {
                        printf( "1\n" ) ; return 0 ;
            }
            memset( num , 0 , sizeof( num ) ) ;
            for ( int i = 0 ; i ++ < n - 2 ; ) INC( i ) ;
            for ( int i = 0 ; i ++ < n - sum - 2 ; ) INC( n - cnt ) ;
            for ( int i = 0 ; i ++ < cnt ; ) {
                        for ( int j = 0 ; j ++ < d[ i ] ; ) DEC( j ) ;
            }
            for ( int i = 0 ; i ++ < n - sum - 2 ; ) DEC( i ) ;
            for ( int i = 0 ; i ++ < p[ 0 ] ; ) {
                        for ( int j = 0 ; j ++ < num[ i ] ; ) ans.multi( p[ i ] ) ;
            }
            ans.print(  ) ;
            return 0 ;
}