Randomly finding a number from the series 1, 2, 3... n to make it m

Using the knowledge of permutation and combination, we can calculate that n bit can represent 2 ^ n cases, for example, 3 bit can represent 8 conditions:
000 001 010 011 100 101 110 111
Now we stipulate that in the n bits, if this bit is 1, then from right to left, how many BITs is this bit: Suppose it is t, then t is selected: for example: in 110, 3 and 2 are selected, and 1 is not selected.
The specific implementation algorithm is as follows:
[Java]
Public static void getNM (int n, int m)
{


Int num = 1 <n;
For (int I = 1; I <= num; I ++)
{
Int sum = 0;
For (int j = I, k = 1; j> 0; j = j> 1, k ++)
{
If (j & 1) = 1)
{
Sum + = k;
}
}

If (sum = m)
{
Int val = 1;
For (int k = I; k> 0; k = k> 1, val ++)
{
If (k & 1) = 1)
{
System. out. print (val + "");
}
}
System. out. println ();
}
}

}