检查数字是否可以表示为2的非零次幂之和

给定一个整数N ，任务是检查N是否可以表示为2的幂之和，其中所有幂> 0，即2 ⁰不能加和。
例子：

Input: N = 10 
Output: 1 
2³ + 2¹ = 10
Input: N = 9 
Output: 0 
 

方法：有两种情况：

1. 当N为偶数时，当幂> 0时，它总是可以表示为2的幂之和。
2. 当N为奇数时，如果总和中不包括2 ^0，则它永远不能表示为2的幂的和。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that return true if n
// can be represented as the sum
// of powers of 2 without using 2^0
bool isSumOfPowersOfTwo(int n)
{
    if (n % 2 == 1)
        return false;
    else
        return true;
}
 
// Driver code
int main()
{
    int n = 10;
    if (isSumOfPowersOfTwo(n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function that return true if n
    // can be represented as the sum
    // of powers of 2 without using 2^0
    static boolean isSumOfPowersOfTwo(int n)
    {
        if (n % 2 == 1)
            return false;
        else
            return true;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        if (isSumOfPowersOfTwo(n))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}

Python3

# Python3 implementation of the approach
 
# Function that return true if n
# can be represented as the sum
# of powers of 2 without using 2^0
def isSumOfPowersOfTwo(n):
    if n % 2 == 1:
        return False
    else:
        return True
 
# Driver code
n = 10
if isSumOfPowersOfTwo(n):
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Shrikant13

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function that return true if n
    // can be represented as the sum
    // of powers of 2 without using 2^0
    static bool isSumOfPowersOfTwo(int n)
    {
        if (n % 2 == 1)
            return false;
        else
            return true;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 10;
        if (isSumOfPowersOfTwo(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

Javascript

// Javascript implementation of the approach
 
// Function that return true if n
// can be represented as the sum
// of powers of 2 without using 2^0
function isSumOfPowersOfTwo(n)
{
    if (n % 2 == 1)
        return false;
    else
        return true;
}
 
// Driver code
var n = 10;
if (isSumOfPowersOfTwo(n))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by noob2000.

输出：

Yes

时间复杂度： O(1)