The beauty of programming: 2.1 calculate the number of 1 in the binary number

Suppose there are n

In the past, logn used the division of two remainder methods to describe the complexity of logn.

There is a logv method where the number of V is 1 is less complex than logn.

 

int Count(int x){int ans = 0;while(x){x &= (x-1);ans++;}return ans;}

Here we use the bitwise Operation X & (x-1) to remove one at a time.

100010001000 & (100010001000-1) = 100010000000

100010000000 & (100010000000-1) = 100000000000

100000000000 & (100000000000-1) = 000000000000

Remove 1 at a time three times in total

 

Then add

for(int i = s; i; i = (i-1)&s)

This loop enumerates all the subsets of S, for example, S = 10101.

10101-> 10100-> 10001-> 10000-> 00101-> 00100-> 00001-> 00000

In this way, all the subsets of S are enumerated.

This will be used when the status is compressed to DP