Seeking the number n, the combination of any number of m

. 1  Import java.util.Scanner;
 2  
. 3  public  class composition Qiuzi problem set 2 
 . 4  {
 . 5      static  void Perm ( int n-, int K)
 . 6      {
 . 7          // int NUM = 0; 
. 8          for ( int I = 0; I <(n-<<. 1); I ++ )
 . 9          {
 10              int KK = I;
 . 11              int NUM = 0 ;
 12 is              the while (KK> 0 ) // KK has been changed
 13 is              {
 14                  KK KK & = (. 1-KK); // meaning of the expression is eliminated after the end of the 1 most, the right to express the value of
 15                  NUM ++ ; // count the number of 1's number
 16              }
 17              IF (NUM == k) // if the number 1 number as num, on the outputs 1 digit
 18 is              {
 . 19                  for ( int J = 0; J <n-; J ++ )
 20 is                      IF (! (I & (1 << J)) = 0 )
 21 is                          of System.out.print (J);
 22 is                  System.out.println ();
 23 is              }
 24          }
 25      }
 26 is      public  static  void main (String [] args) 
 27      {
28          // the TODO method of automatically generating stubs 
29          Scanner SC = new new Scanner (the System.in);
 30          int n-= sc.nextInt (); // number represents n-
 31 is          int K = sc.nextInt (); // represents The number m
 32          Perm (n-, K);
 33 is      }
 34 is  
35 }

It should be noted that point, the code number of the solution is a combination of an arbitrary number n m.

kk & (kk-1), the meaning of which is to remove the end of the expression 1, such as: & 1011 (1011-1) = 1011 = 1010 & 1010 

1010&(1010-1)  = 1010&1001 =1000