检查原始数组总和是奇数还是偶数使用数组的按位与

给定一个表示数组大小的整数N和数组所有元素的按位与(K) 。任务是确定元素的总和是奇数还是偶数或无法确定。

例子：

Input: N = 1, K = 11
Output: Odd
Explanation: As there is only one element in the array, the element itself is 11.

Input: N = 1, K = 2
Output: Even
Explanation: As there is only one element in the array. the element itself is 2.

Input: N = 5, K = 4
Output: Impossible

方法：该问题的解决方案基于以下观察：

Observation: 
 

• We know that for bitwise AND to be 1, all the bits must be 1. 
   
• So if the given LSB of given bitwise AND of Array is 1, it must mean that LSB of all the elements of Array must have been 1.
• On the other hand, if the LSB of given bitwise AND of Array is 0, it can mean that either all or some of the elements of Array has 0 in their LSB.

Conclusion: Therefore using above observation, it can be said that: 
 

• For bitwise AND to be odd, all the elements of the array must be odd, as LSB of odd numbers is always 1.
• But if the bitwise AND is even, then nothing can be said about the parity (odd or even) of all the elements.

请按照以下步骤解决问题：

• 如果K的LSB为 1 ，则清楚地证明每个数字的 LSB 为 1 即每个数字都是奇数，因为如果任何单个数字的 LSB 为 0，则K的 LSB 将是 0。
  • 如果N为奇数，则N个奇数之和总是奇数。
  • 和是偶数，如果N是偶数。
• 如果K的LSB为 0，则无法确定总和，除非N 为 1的情况，即N为1时总和取决于K的奇偶性。在另一种情况下，由于无法确定偶数和奇数的计数，因此无法找到。

下面是上述方法的实现。

C++

// C++ code to implement the above approach
#include 
using namespace std;
 
// Function to find whether the sum is even,
// odd or impossible to find.
void findtotalsum(int n, int a)
{
    // Separate case when N is 1
    // as it depend on parity of n
    if (n == 1) {
        if (a % 2 == 0)
            cout << "Even" << endl;
        else
            cout << "Odd" << endl;
    }
 
    // Check if a is odd
    else if (a % 2 != 0) {
 
        // Checking whether n is odd or even
        if (n % 2 == 0)
            cout << "Even" << endl;
        else
            cout << "Odd" << endl;
    }
    else {
 
        // When no condition applies
        // its impossible to find sum
        cout << "Impossible" << endl;
    }
}
 
// Driver code
int main()
{
    long int N, K;
    N = 5, K = 4;
    findtotalsum(N, K);
    return 0;
}

Java

// Java code to minimize number of rotations
import java.util.*;
 
class GFG {
 
  // Function to find whether the sum is even,
  // odd or impossible to find.
  static void findtotalsum(int n, int a)
  {
 
    // Separate case when N is 1
    // as it depend on parity of n
    if (n == 1) {
      if (a % 2 == 0)
        System.out.println("Even");
      else
        System.out.println("Odd");
    }
 
    // Check if a is odd
    else if (a % 2 != 0) {
 
      // Checking whether n is odd or even
      if (n % 2 == 0)
        System.out.println("Even");
      else
        System.out.println("Odd");
    }
    else {
 
      // When no condition applies
      // its impossible to find sum
      System.out.println("Impossible");
    }
  }
 
  // Driver code
  public static void main (String[] args) {
    int N = 5, K = 4;
    findtotalsum(N, K);
  }
}
 
// This code i contributed by hrithikgarg03188.

Python3

# Python program for the above approach
 
# Function to find whether the sum is even,
# odd or impossible to find.
def findtotalsum(n, a) :
 
    # Separate case when N is 1
    # as it depend on parity of n
    if (n == 1) :
        if (a % 2 == 0) :
            print("Even")
        else :
            print( "Odd" )
     
 
    # Check if a is odd
    elif (a % 2 != 0) :
 
        # Checking whether n is odd or even
        if (n % 2 == 0) :
            print("Even")
        else :
            print( "Odd" )
     
    else :
 
        # When no condition applies
        # its impossible to find sum
        print( "Impossible")
     
# Driver code
 
N = 5
K = 4
findtotalsum(N, K)
 
# This code is contributed by sanjoy_62.

C#

// C# code to minimize number of rotations
using System;
 
class GFG {
 
  // Function to find whether the sum is even,
  // odd or impossible to find.
  static void findtotalsum(int n, int a)
  {
 
    // Separate case when N is 1
    // as it depend on parity of n
    if (n == 1) {
      if (a % 2 == 0)
        Console.WriteLine("Even");
      else
        Console.WriteLine("Odd");
    }
 
    // Check if a is odd
    else if (a % 2 != 0) {
 
      // Checking whether n is odd or even
      if (n % 2 == 0)
        Console.WriteLine("Even");
      else
        Console.WriteLine("Odd");
    }
    else {
 
      // When no condition applies
      // its impossible to find sum
      Console.WriteLine("Impossible");
    }
  }
 
  // Driver code
  public static void Main () {
    int N = 5, K = 4;
    findtotalsum(N, K);
  }
}
 
// This code i contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code to implement the above approach
 
    // Function to find whether the sum is even,
    // odd or impossible to find.
    const findtotalsum = (n, a) => {
     
        // Separate case when N is 1
        // as it depend on parity of n
        if (n == 1) {
            if (a % 2 == 0)
                document.write("Even<br/>");
            else
                document.write("Odd<br/>");
        }
 
        // Check if a is odd
        else if (a % 2 != 0) {
 
            // Checking whether n is odd or even
            if (n % 2 == 0)
                document.write("Even<br/>");
            else
                document.write("Odd<br/>");
        }
        else {
 
            // When no condition applies
            // its impossible to find sum
            document.write("Impossible<br/>");
        }
    }
 
    // Driver code
    let N, K;
    N = 5, K = 4;
    findtotalsum(N, K);
 
// This code is contributed by rakeshsahni
 

输出

Impossible

时间复杂度： O(1)
辅助空间： O(1)