检查给定的数字是否是 d 的幂，其中 d 是 2 的幂
原文:https://www . geesforgeks . org/check-给定-number-power-d-d-power-2/
给定一个整数 n，求它是否是 d 的幂，其中 d 本身就是 2 的幂。 例:
Input : n = 256, d = 16
Output : Yes
Input : n = 32, d = 16
Output : No
方法 1 取给定数在基数 d 上的对数，如果得到一个整数，那么这个数就是 d 的幂。 方法 2 继续将这个数除以 d，即迭代做 n = n/d。在任何迭代中，如果 n%d 变成非零且 n 不是 1，则 n 不是 d 的幂，否则 n 是 d 的幂。 方法 3(按位) 如果满足以下条件，则数字 n 是 d 的幂。 a)在 n 的二进制表示中只有一个位集(注意:d 是 2 的幂) b)在(唯一)设置位之前的零位计数是 log 的倍数 2 (d)。 例如:对于 n = 16 (10000)和 d = 4，16 是 4 的幂，因为只有一个位设置，并且在设置位是 4 之前计数为 0，4 是 log 的倍数 2 (4)。
C++
// CPP program to find if a number is power
// of d where d is power of 2.
#include<stdio.h>
unsigned int Log2n(unsigned int n)
{
return (n > 1)? 1 + Log2n(n/2): 0;
}
bool isPowerOfd(unsigned int n, unsigned int d)
{
int count = 0;
/* Check if there is only one bit set in n*/
if (n && !(n&(n-1)) )
{
/* count 0 bits before set bit */
while (n > 1)
{
n >>= 1;
count += 1;
}
/* If count is a multiple of log2(d)
then return true else false*/
return (count%(Log2n(d)) == 0);
}
/* If there are more than 1 bit set
then n is not a power of 4*/
return false;
}
/* Driver program to test above function*/
int main()
{
int n = 64, d = 8;
if (isPowerOfd(n, d))
printf("%d is a power of %d", n, d);
else
printf("%d is not a power of %d", n, d);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find if
// a number is power of d
// where d is power of 2.
class GFG
{
static int Log2n(int n)
{
return (n > 1)? 1 +
Log2n(n / 2): 0;
}
static boolean isPowerOfd(int n,
int d)
{
int count = 0;
/* Check if there is
only one bit set in n*/
if (n > 0 && (n &
(n - 1)) == 0)
{
/* count 0 bits
before set bit */
while (n > 1)
{
n >>= 1;
count += 1;
}
/* If count is a multiple
of log2(d) then return
true else false*/
return (count %
(Log2n(d)) == 0);
}
/* If there are more
than 1 bit set then
n is not a power of 4*/
return false;
}
// Driver Code
public static void main(String[] args)
{
int n = 64, d = 8;
if (isPowerOfd(n, d))
System.out.println(n +
" is a power of " + d);
else
System.out.println(n +
" is not a power of " + d);
}
}
// This code is contributed by mits
Python 3
# Python3 program to find if a number
# is power of d where d is power of 2.
def Log2n(n):
return (1 + Log2n(n / 2)) if (n > 1) else 0;
def isPowerOfd(n, d):
count = 0;
# Check if there is only
# one bit set in n
if (n and (n & (n - 1))==0):
# count 0 bits
# before set bit
while (n > 1):
n >>= 1;
count += 1;
# If count is a multiple of log2(d)
# then return true else false
return (count%(Log2n(d)) == 0);
# If there are more than 1 bit set
# then n is not a power of 4
return False;
# Driver Code
n = 64;
d = 8;
if (isPowerOfd(n, d)):
print(n,"is a power of",d);
else:
print(n,"is not a power of",d);
# This code is contributed by mits
C
// C# program to find if
// a number is power of d
// where d is power of 2.
using System;
class GFG
{
static int Log2n(int n)
{
return (n > 1)? 1 +
Log2n(n / 2): 0;
}
static bool isPowerOfd(int n,
int d)
{
int count = 0;
/* Check if there is
only one bit set in n*/
if (n > 0 && (n & (n - 1)) == 0)
{
/* count 0 bits
before set bit */
while (n > 1)
{
n >>= 1;
count += 1;
}
/* If count is a multiple
of log2(d) then return
true else false*/
return (count % (Log2n(d)) == 0);
}
/* If there are more than
1 bit set then n is not
a power of 4*/
return false;
}
// Driver Code
static void Main()
{
int n = 64, d = 8;
if (isPowerOfd(n, d))
Console.WriteLine("{0} is a " +
"power of {1}",
n, d);
else
Console.WriteLine("{0} is not a"+
" power of {1}",
n, d);
}
// This code is contributed by mits
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find if a number
// is power of d where d is power of 2.
function Log2n($n)
{
return ($n > 1)? 1 +
Log2n($n / 2): 0;
}
function isPowerOfd($n, $d)
{
$count = 0;
// Check if there is only
// one bit set in n
if ($n && !($n & ($n - 1)))
{
// count 0 bits
// before set bit
while ($n > 1)
{
$n >>= 1;
$count += 1;
}
/* If count is a multiple of log2(d)
then return true else false*/
return ($count%(Log2n($d)) == 0);
}
/* If there are more than 1 bit set
then n is not a power of 4*/
return false;
}
// Driver Code
$n = 64;
$d = 8;
if (isPowerOfd($n, $d))
echo $n," ","is a power of ", $d;
else
echo $n," ","is not a power of ", $d;
// This code is contributed by m_kit
?>
java 描述语言
<script>
// Javascript program to find if
// a number is power of d
// where d is power of 2
function Log2n(n)
{
return (n > 1) ? 1 +
Log2n(n / 2) : 0;
}
// Function to count the number
// of ways to paint N * 3 grid
// based on given conditions
function isPowerOfd(n, d)
{
var count = 0;
/* Check if there is
only one bit set in n*/
if (n > 0 && (n & (n - 1)) == 0)
{
/* count 0 bits
before set bit */
while (n > 1)
{
n >>= 1;
count += 1;
}
/* If count is a multiple
of log2(d) then return
true else false*/
return (count % (Log2n(d)) == 0);
}
/* If there are more
than 1 bit set then
n is not a power of 4*/
return false;
}
// Driver code
var n = 64, d = 8;
if (isPowerOfd(n, d))
document.write(n +
" is a power of " + d);
else
document.write(n +
" is not a power of " + d);
// This code is contributed by Khushboogoyal499
</script>
Output:
64 is a power of 8
时间复杂度: O(log 2 n)
辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
