给定一个数字,检查它是否为8的幂。
例子 :
Input : n = 64
Output : Yes
Input : 75
Output : No
第一个解决方案
我们计算数字的log8(n)(如果它是整数),则n为8的幂。我们使用trunc(n)函数为双精度值找到最接近的整数。
C++
// C++ program to check if a number is power of 8
#include
#include
using namespace std;
/* function to check if power of 8 */
bool checkPowerof8(int n)
{
/* calculate log8(n) */
double i = log(n) / log(8);
/* check if i is an integer or not */
return (i - trunc(i) < 0.000001);
}
/* driver function */
int main()
{
int n = 65;
checkPowerof8(n) ? cout << "Yes" : cout << "No";
return 0;
}
Java
// Java program to check if
// a number is power of 8
class GFG {
// function to check
// if power of 8
static boolean checkPowerof8(int n)
{
/* calculate log8(n) */
double i = Math.log(n) / Math.log(8);
/* check if i is an integer or not */
return (i - Math.floor(i) < 0.000001);
}
// Driver Code
public static void main(String args[])
{
int n = 65;
if (checkPowerof8(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Sam007
Python3
# Python3 program to check
# if a number is power of 8
from math import log,trunc
# function to check if power of 8
def checkPowerof8(n):
# calculate log8(n)
i = log(n, 8)
# check if i is an integer or not
return (i - trunc(i) < 0.000001);
# Driver Code
n = 65
if checkPowerof8(n):
print("Yes")
else:
print("No")
# This code is contributed by Mohit Kumar
C#
// C# program to check if
// a number is power of 8
using System;
class GFG {
// function to check
// if power of 8
static bool checkPowerof8(int n)
{
// calculate log8(n) */
double i = Math.Log(n) / Math.Log(8);
// check if i is an integer or not */
return (i - Math.Floor(i) < 0.000001);
}
// Driver Code
static public void Main()
{
int n = 65;
if (checkPowerof8(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by akt_mit
PHP
Javascript
C++
// C++ program to check if a number is power of 8
// using bit mask.
#include
using namespace std;
/*function to check if power of 8*/
bool checkPowerof8(int n)
{
return (n && !(n & (n - 1)) && !(n & 0xB6DB6DB6));
}
/*driver function*/
int main()
{
int n = 65;
checkPowerof8(n) ? cout << "Yes" : cout << "No";
return 0;
}
Java
// Java program to check if a
// number is power of 8 using
// bit mask.
import java.util.*;
class GFG{
// function to check if
// power of 8
static boolean checkPowerof8(int n)
{
return (n > 0 && (n & (n - 1)) > 0 &&
(n & 0xB6DB6DB6) > 0);
}
// Driver code
public static void main(String[] args)
{
int n = 65;
if (checkPowerof8(n) == true)
System.out.print("Yes" );
else
System.out.print("No");
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to check if a number
# is power of 8
# function to check if power of 8
def checkPowerof8(n):
return (n and not (n & (n - 1)) and
not (n & 0xB6DB6DB6))
# Driver Code
if __name__ == "__main__":
n = 65
if checkPowerof8(n):
print ("Yes")
else:
print ("No")
# This code is contributed by ita_c
C#
// C# program to check if a
// number is power of 8 using
// bit mask.
using System;
class GFG{
// Function to check if
// power of 8
static bool checkPowerof8(int n)
{
return (n > 0 && (n & (n - 1)) > 0 &&
(n & 0xB6DB6DB6) > 0);
}
// Driver code
static public void Main()
{
int n = 65;
if (checkPowerof8(n) == true)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by avanitrachhadiya2155
PHP
Javascript
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to check if n is power of 8
bool checkPowerof8(int n)
{
// Variable i will denote the bit
// that we are currently at
int i = 0;
unsigned long long l = 1;
while (i <= 63) {
l <<= i;
// If only set bit in n
// is at position i
if (l == n)
return true;
// Get to next valid bit position
i += 3;
l = 1;
}
return false;
}
// Driver code
int main()
{
int n = 65;
if (checkPowerof8(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to check if n is power of 8
static boolean checkPowerof8(int n)
{
// Variable i will denote the bit
// that we are currently at
int i = 0;
long l = 1;
while (i <= 63)
{
l <<= i;
// If only set bit in n
// is at position i
if (l == n)
return true;
// Get to next valid bit position
i += 3;
l = 1;
}
return false;
}
// Driver code
public static void main (String[] args)
{
int n = 65;
if (checkPowerof8(n))
System.out.println ("Yes");
else
System.out.println( "No");
}
}
// This code is contributed by Tushil.
Python3
# Python3 implementation of the approach
# Function to check if n is power of 8
def checkPowerof8(n):
# Variable i will denote the bit
# that we are currently at
i = 0
l = 1
while (i <= 63):
l <<= i
# If only set bit in n
# is at position i
if (l == n):
return True
# Get to next valid bit position
i += 3
l = 1
return False
# Driver code
if __name__ == '__main__':
n = 65
if (checkPowerof8(n)):
print("Yes")
else:
print("No")
# This code is contributed by math_lover
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to check if n is power of 8
static bool checkPowerof8(int n)
{
// Variable i will denote the bit
// that we are currently at
int i = 0;
long l = 1;
while (i <= 63)
{
l <<= i;
// If only set bit in n
// is at position i
if (l == n)
return true;
// Get to next valid bit position
i += 3;
l = 1;
}
return false;
}
// Driver code
static public void Main ()
{
int n = 65;
if (checkPowerof8(n))
Console.WriteLine("Yes");
else
Console.WriteLine( "No");
}
}
// This code is contributed by ajit.
Javascript
输出 :
No
第二解决方案
如果满足以下条件,则数字是8的幂。
- 该数字是2的幂。如果数字只有一个置1的位,即为n且n-1为0,则该数字为2的幂。
- 该数字仅在位置0或3或6或…上设置了一位。 30 [对于32位数字]。要检查其设置位的位置,我们可以使用掩码(0xB6DB6DB6) 16 =(10110110110110110110110110110110110) 2 。
以下是上述想法的实现。
C++
// C++ program to check if a number is power of 8
// using bit mask.
#include
using namespace std;
/*function to check if power of 8*/
bool checkPowerof8(int n)
{
return (n && !(n & (n - 1)) && !(n & 0xB6DB6DB6));
}
/*driver function*/
int main()
{
int n = 65;
checkPowerof8(n) ? cout << "Yes" : cout << "No";
return 0;
}
Java
// Java program to check if a
// number is power of 8 using
// bit mask.
import java.util.*;
class GFG{
// function to check if
// power of 8
static boolean checkPowerof8(int n)
{
return (n > 0 && (n & (n - 1)) > 0 &&
(n & 0xB6DB6DB6) > 0);
}
// Driver code
public static void main(String[] args)
{
int n = 65;
if (checkPowerof8(n) == true)
System.out.print("Yes" );
else
System.out.print("No");
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to check if a number
# is power of 8
# function to check if power of 8
def checkPowerof8(n):
return (n and not (n & (n - 1)) and
not (n & 0xB6DB6DB6))
# Driver Code
if __name__ == "__main__":
n = 65
if checkPowerof8(n):
print ("Yes")
else:
print ("No")
# This code is contributed by ita_c
C#
// C# program to check if a
// number is power of 8 using
// bit mask.
using System;
class GFG{
// Function to check if
// power of 8
static bool checkPowerof8(int n)
{
return (n > 0 && (n & (n - 1)) > 0 &&
(n & 0xB6DB6DB6) > 0);
}
// Driver code
static public void Main()
{
int n = 65;
if (checkPowerof8(n) == true)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by avanitrachhadiya2155
的PHP
Java脚本
输出 :
No
一个简单的观察结果就是,如果一个数字是8的幂,那么它只有一个1位的集合,并且该位位于位置1、4、7、10,…上。
因此,我们可以仅检查数字中设置的唯一位是否在这些位置之一,那么它是否为8的幂。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to check if n is power of 8
bool checkPowerof8(int n)
{
// Variable i will denote the bit
// that we are currently at
int i = 0;
unsigned long long l = 1;
while (i <= 63) {
l <<= i;
// If only set bit in n
// is at position i
if (l == n)
return true;
// Get to next valid bit position
i += 3;
l = 1;
}
return false;
}
// Driver code
int main()
{
int n = 65;
if (checkPowerof8(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to check if n is power of 8
static boolean checkPowerof8(int n)
{
// Variable i will denote the bit
// that we are currently at
int i = 0;
long l = 1;
while (i <= 63)
{
l <<= i;
// If only set bit in n
// is at position i
if (l == n)
return true;
// Get to next valid bit position
i += 3;
l = 1;
}
return false;
}
// Driver code
public static void main (String[] args)
{
int n = 65;
if (checkPowerof8(n))
System.out.println ("Yes");
else
System.out.println( "No");
}
}
// This code is contributed by Tushil.
Python3
# Python3 implementation of the approach
# Function to check if n is power of 8
def checkPowerof8(n):
# Variable i will denote the bit
# that we are currently at
i = 0
l = 1
while (i <= 63):
l <<= i
# If only set bit in n
# is at position i
if (l == n):
return True
# Get to next valid bit position
i += 3
l = 1
return False
# Driver code
if __name__ == '__main__':
n = 65
if (checkPowerof8(n)):
print("Yes")
else:
print("No")
# This code is contributed by math_lover
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to check if n is power of 8
static bool checkPowerof8(int n)
{
// Variable i will denote the bit
// that we are currently at
int i = 0;
long l = 1;
while (i <= 63)
{
l <<= i;
// If only set bit in n
// is at position i
if (l == n)
return true;
// Get to next valid bit position
i += 3;
l = 1;
}
return false;
}
// Driver code
static public void Main ()
{
int n = 65;
if (checkPowerof8(n))
Console.WriteLine("Yes");
else
Console.WriteLine( "No");
}
}
// This code is contributed by ajit.
Java脚本
输出 :
No