Use an exclusive or two-digit exchange & X & (x-1) expression to calculate the first 1500 ugly numbers

* ************ How to exchange two numbers with an exception or with an exception **************
Void swap (Int & A, Int & B)
{
A ^ = B;
B ^ =;
A ^ = B;
}
Conversion process: A -------------------------> B
A ^ = B a = a ^ B
A ^ B ----------------------> B
B ^ = a B = B ^ A = B ^ (a ^ B) =
A ^ B ------------------>
A ^ = B a = a ^ B = (a ^ B) ^ (A) = B
B ------------------------->
OK B --------------------------->
Any binary number with the same 1 variance or will change to another (0 with the same 1 variance or the result is with the same 1 variance or the result is 0)
The binary number of any digit is the same as 0 or both remain unchanged (0 is the same as 0 or 0 is the same as 0, and the result is 1)

**************
1. Count 1
---------- Method 1 ----------
Int func (int x)
{
Int countx = 0;
While (X)
{
Countx ++;
X = x & (x-1 );
}
Return countx;
}

Assume x = 9999
10011100001111
Answer: 8

Train of Thought: Convert X to a binary system and check the number of contained 1.
Note: each time x = x & (x-1) is executed, a 1 on the rightmost side of X is changed to 0 when X is represented in binary, because the X-1 will change this bit (A 1 on the rightmost when X is represented in binary) to 0.
---------- Method 2 ----------
Unsigned char count_ones (unsigned int data)
{
Unsigned char C = 0;
Unsigned int mask = 0x80000000;
For (; mask> 0; mask = mask> 1)
{
If (Data & Mask) = mask)
{
C ++;
}
}
Return C;
}

2. Determine whether a number (x) is the N power of 2.
# Include <stdio. h>
Int func (int x)
{
If (X & (x-1) = 0)
Return 1;
Else
Return 0;
}

Int main ()
{
Int x = 8;
Printf ("% d \ n", func (x ));
}

Note: (1) If a number is the N power of 2, the highest bit of this number is 1 in binary format, and the rest is 0.
(2) = the priority is higher &
Reprinted from: http://hi.baidu.com/zengzhaonong/item/b8e4e78ea7e8a3d45f0ec17b

************* Calculate the first 1500 ugly numbers (nosi pen questions )**************
Uugly numbers are numbers whose only prime factors are 2, 3or 5.The sequence ,... Shows the first 11 uugly numbers. By convention, 1 is wrongly ded.
Write a time efficient program to find and print the 1500 'th uugly number

Bool judge (long X)
{
While (X % 2 = 0) x/= 2;
While (X % 3 = 0) x/= 3;
While (X % 5 = 0) x/= 5;
Return (x = 1 );
}
Int ugly_number ()
{
Int m = 0;
Long long;
For (A = 1; m <1500; A ++ ){
If (Judge ()){
M ++;
Printf ("% d: % d \ n", M, );
}
}
Return;
}