Question

How can I find perfect power of two between two numbers? Sample input: 0 and 10 Output: 2, 4, 8

Answer 1

Well the interesting part is "How do I get the greatest power of 2 that is less than or equal to my upper bound" and the same for the lowest power of 2 that is greater or equal to the lower bound.

而且,这很容易在没有住所的情况下进行。未签署32个借项号:

floor(x):   ; floor power of 2
    x = x | (x >> 1)
    x = x | (x >> 2)
    x = x | (x >> 4)
    x = x | (x >> 8)
    x = x | (x >> 16)
    return x - (x >> 1)

ceil(x):   ; ceiling power of 2
    x = x - 1
    x = x | (x >> 1)
    x = x | (x >> 2)
    x = x | (x >> 4)
    x = x | (x >> 8)
    x = x | (x >> 16)
    return x + 1

你们获得了生产数字的通道,但 well得好。

Answer 2

选择第一组最高点为1级,即从最低点算起。之后,第二位数中位数为一,即处于正轨状态。 2^x+1, 2^x+2......, 2^y 是你重新寻找的人数。

Answer 3

您可使用数字和产出的双位代表,在只设定一个比值的中间点之间:

0  = 00000000
10 = 00001010

=>

     00000001 (1)
     00000010 (2)
     00000100 (4)
     00001000 (8)

因此,你的问题被缩小,以找到比最低点更长的头一种权力,然后在你比最高点更低时转移。另一种做法是,除最高值外,将所有设定的限额固定在最高值之内,然后在你超过最低值时转移权利。

Answer 4

How many times can you bit-shift before getting to 0?

Answer 5

步骤:

Let s say n1 = start_of_range, n2 = end_of_range.
Find out how many bits are needed to represent n1. Call it b.
Now 2**b will be the next power of two after n1.
It should be easy to calculate all the power of twos till n2 from this.

Sample python Code:

#!/usr/bin/python
def get_bits(n):
    b = 0
    while n:
        n = n / 2
        b += 1
    return b

def get_power_of_two_in_range(n1, n2):
    b = get_bits(n1)
    x = 2**b
    res = []
    while x < n2:
        res.append(x)
        x = x * 2
    return res

def main():
    print get_power_of_two_in_range(0, 10)

if __name__ ==  __main__ :
    main()

Answer 6

int first=0; //**first number which is a complete power of 2 in the range!**
 for(i=low;i<high;i++){
  if(2i==(i^(i-1)+1)){
   first=i;
   break;
  }
 }

while(first!=0){
 print first;
 first*=2;
 if(first>high){
  break;
}
}

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