hdu1521-排列组合

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http://acm.hdu.edu.cn/showproblem.php?pid=1521

指数型母函数

G（x) = ( 1 + x / 1! + (x^2)/(2!) + .....+ (x ^ n1 ) / （n1!) ) * （( 1 + x / 1! + (x^2)/(2!) + .....+ (x ^ n２ ) / （n２!)*.............*(( 1 + x / 1! + (x^2)/(2!) + .....+ (x ^ nk) / （nk!) )

求x^m的系数转换成temp / （ m ! ) ;

然后求解时，temp * （m! ) 就是x ^ m的系数

#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 ;double fac[ ] = { 1 , 1 , 2 , 6 , 24 , 120 , 720 , 5040 , 40320 , 362880 , 3628800 , 39916800 } ;int main(){int n , m ;double a[ 11 ] , b[ 11 ] , num1[ 11 ], num2[ 11 ];while( scanf( "%d%d" , &n , &m ) != EOF ){for( int i = 0 ; i < n ; ++i ){scanf( "%lf" , &a[ i ] ) ;}for( int i = 0 ; i <= m ; ++i )num1[ i ] = num2[ i ] = 0.0 ;for( int i = 0 ; i <= a[ 0 ] ; ++i ){num1[ i ] = 1.0 / fac[ i ] ;}for( int i = 1 ; i < n ; ++i ){for( int j = 0 ; j <= m ; ++j ){for( int k = 0 ; k <= a[ i ] && k + j <= m ; ++k ){num2[ k + j ] += ( num1[ j ] / fac[ k ] ) ;}}for( int j = 0 ; j <= m ; ++j ){num1[ j ] = num2[ j ] ;num2[ j ] = 0 ; }}printf( "%.lf\n" , num1[ m ] * fac[ m ] ) ;}    return 0;}