How many integers are in the binary notation of 1?

Reprinted from:Http://www.cnblogs.com/python27/.

Question: enter an integer to determine the number of 1 in the binary representation of the positive number? For example, if the input integer 12 is converted to a binary value of 1100, there are two 1 in total, SO 2 should be output.

 

Analysis 1: We can consider how to determine whether each bit is 1 from right to left? We have this number and integer 1 (01) done and operate. Because 1 except the last digit, all the other parts are 0, if the last digit of the integer is 1, the return result is 1. If the last 1 of an integer is 0, the return result is 0. Then, let the integer and 2 (10) perform the operation, then, we can determine whether the second-to-last integer is 1. Then, let the integer and the fourth (100) perform an operation to determine whether the third-to-last integer is 1. and so on, we can determine the number of all 1 integers, and the number of operations with integers is 1, 2, 4, 8 ...... we can perform the 1-bit left shift operation. Based on this idea, we can get the following:Code:

 1 # Include <iostream>
 2 # Include < String >
 3   Using  Namespace STD;
 4  
 5   Int Numbersof1s ( Int Number)
 6 {
 7       Int CNT = 0 ;
 8 Unsigned Int Flag = 1 ;
9       While (FLAG)
 10 {
 11           If (Flag & number)
 12 CNT ++;
 13  
 14 Flag = Flag < 1 ;
 15 }
 16       Return CNT;
17 }
 18  
 19   Int Main ()
 20 {
 21 Cout < "  Enter a number:  " <Endl;
 22       Int I;
 23 Cin> I;
 24 Cout < "  The numbers of 1 s in your number is:  " 
 25 <Numbersof1s (I) <Endl;
 26       Return   0 ;
 27 }

The running result is as follows:

Analysis 2: If an integer is not 0, at least one of its binary forms is 1. If we subtract 1 from a binary number, then the rightmost 1 of the binary number will be changed to 0, and the 0 after this 1 will be changed to 1. take 1100 as an example. The rightmost 1 is the third digit of the right number. After this number is subtracted from 1, it becomes 1011. If we calculate the obtained number and original number, the first 1 on the right and the number after it are both 0, that is, 1100 & 1011 = 1000, that is, after a number is subtracted from 1, and the operation is performed on the original number, the rightmost 1 and the number of 1 will be eliminated. Then we can perform a similar loop as many times. Based on this idea, we have the following code:

1 # Include <iostream>
 2 # Include < String >
 3   Using   Namespace STD;
 4  
 5   Int Numbersof1s ( Int Number)
 6 {
 7       Int CNT = 0 ;
 8       While (Number)
 9 {
 10 CNT ++;
 11 Number = Number & (number- 1 );
 12 }
 13  
 14       Return CNT;
15 }
 16  
 17   Int Main ()
 18 {
 19 Cout < "  Enter a number:  " <Endl;
 20       Int I;
 21 Cin> I;
 22 Cout < "  The numbers of 1 s in your number is:  " 
 23 <Numbersof1s (I) <Endl;
 24       Return   0 ;
 25 }

The running result is as follows:

Concerning the shift operation and multiplication and division operation: In Analysis 1, we can say that we can continue to perform a one-digit operation to get the values of 1, 2, 4, 8, and 16 in decimal format. Some may say, why do we not perform the X 2 operation on the flag? Because the efficiency of the shift operation is much higher than the multiplication and division operation, it is not just a question. In other programming, we should also try to replace the multiplication and division operations with the shift operation.