给定一个整数n，找出它是否是d的幂，其中d本身是2的幂。 例如：
null
Input : n = 256, d = 16Output : YesInput : n = 32, d = 16Output : No
方法1 以给定的数字为基数d，如果我们得到一个整数，那么这个数字就是d的幂。 方法2 继续用d除以这个数，也就是说，迭代地做n=n/d。在任何迭代中，如果n%d变为非零且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） n的二进制表示中只有一位（注：d是2的幂） b） 在（唯一）设置位之前的零位计数是对数的倍数 2. （d） 。 例如：对于n=16（10000）和d=4，16是4的幂，因为只设置了一个位，并且在设置的位为4之前计数为0，这是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
// 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 |
Python3
# 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 } |
PHP
<?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 ?> |
Javascript
<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> |
输出：
64 is a power of 8
时间复杂性： O（原木） 2. n）
辅助空间： O（1）
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