查找给定范围内具有至少一个奇数除数的元素

给定两个整数N和M ，任务是打印范围[N，M]中具有至少一个奇数除数的所有元素。
例子：

Input: N = 2, M = 10 
Output: 3 5 6 7 9 10 
Explanation: 
3, 6 have an odd divisor 3 
5, 10 have an odd divisor 5 
7 have an odd divisor 7 
9 have two odd divisors 3, 9
Input: N = 15, M = 20 
Output: 15 17 18 19 20 
 

天真的方法：
最简单的方法是循环遍历[1，N]范围内的所有数字，并针对每个元素检查其是否具有奇数除数。
时间复杂度： O(N ^3/2 )
高效方法：
为了优化上述方法，我们需要注意以下细节：

• 2的幂的任何数字都没有奇数除数。
• 所有其他元素将至少具有一个奇数除数

Illustration: 
In the range [3, 10], the elements which have at least one odd divisors are {3, 5, 6, 7, 9, 10} and {4, 8} does not contain any odd divisors. 
 

请按照以下步骤解决问题：

• 遍历范围[N，M]并检查每个元素的设置位是否设置为前一个数字，即检查i＆(i – 1)是否等于0 。
• 如果是这样，则该数字为2的幂，不予考虑。否则，请打印该数字，因为它不是2的幂且具有至少一个奇数除数。

下面是上述方法的实现。

C++

// C++ program to print all numbers
// with least one odd factor in the
// given range
#include 
using namespace std;
 
// Function to prints all numbers
// with at least one odd divisor
int printOddFactorNumber(int n,
                         int m)
{
    for (int i = n; i <= m; i++) {
 
        // Check if the number is
        // not a power of two
        if ((i > 0)
            && ((i & (i - 1)) != 0))
 
            cout << i << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 2, M = 10;
    printOddFactorNumber(N, M);
    return 0;
}

Java

// Java program to print all numbers
// with least one odd factor in the
// given range
class GFG{
 
// Function to prints all numbers
// with at least one odd divisor
static int printOddFactorNumber(int n,
                                int m)
{
    for (int i = n; i <= m; i++)
    {
 
        // Check if the number is
        // not a power of two
        if ((i > 0) && ((i & (i - 1)) != 0))
 
            System.out.print(i + " ");
    }
    return 0;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2, M = 10;
    printOddFactorNumber(N, M);
}
}
 
// This code is contributed
// by shivanisinghss2110

Python3

# Python3 program to print all numbers
# with least one odd factor in the
# given range
 
# Function to prints all numbers
# with at least one odd divisor
def printOddFactorNumber(n, m):
     
    for i in range(n, m + 1):
 
        # Check if the number is
        # not a power of two
        if ((i > 0) and ((i & (i - 1)) != 0)):
            print(i, end = " ")
             
# Driver Code
N = 2
M = 10
 
printOddFactorNumber(N, M)
 
# This code is contributed by Vishal Maurya

C#

// C# program to print all numbers
// with least one odd factor in the
// given range
using System;
class GFG{
 
// Function to prints all numbers
// with at least one odd divisor
static int printOddFactorNumber(int n,
                                int m)
{
    for (int i = n; i <= m; i++)
    {
 
        // Check if the number is
        // not a power of two
        if ((i > 0) && ((i & (i - 1)) != 0))
 
            Console.Write(i + " ");
    }
    return 0;
}
 
// Driver Code
public static void Main()
{
    int N = 2, M = 10;
    printOddFactorNumber(N, M);
}
}
 
// This code is contributed
// by Code_Mech

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to prints all numbers
// with at least one odd divisor
function printOddFactorNumber(n, m)
{
    for (let i = n; i <= m; i++)
    {
   
        // Check if the number is
        // not a power of two
        if ((i > 0) && ((i & (i - 1)) != 0))
   
            document.write(i + " ");
    }
    return 0;
}
 
// Driver code
    let N = 2, M = 10;
    printOddFactorNumber(N, M);
 
// This code is contributed by susmitakundugoaldanga.

输出：

3 5 6 7 9 10

时间复杂度： O(N)
辅助空间： O(1)