检查一个Array是否可以通过反复拆分和合并使为0

给定一个包含N 个元素的数组arr[] ，任务是找到给定的操作是否可以使给定数组的所有元素都为 0。在这个数组上只能执行两种类型的操作：

• 将元素 B 拆分为 2 个元素 C 和 D，使得B = C + D 。
• 将 2 个元素 P 和 Q 合并为一个元素 R，使得R = P^Q即(P 和 Q 的异或)。

例子：

Input:  arr = [9, 17] 
Output: Yes 
Explanation: Following is one possible sequence of operations –   
1) Merge i.e 9 XOR 17 = 24    
2) Split 24 into two parts each of size 12   
3) Merge i.e 12 XOR 12 = 0    
As there is only 1 element i.e 0. So it is possible.

Input:  arr = [1] 
Output: No 
Explanation: There is no possible way to make it 0.

方法 ：

1. 如果数组中的任何元素是偶数，则可以将其设为 0。将该元素拆分为 arr[i]/2 和 arr[i]/2 的两个相等部分。两个相等数的异或为零。因此，该策略使元素为 0。
2. 如果任何元素是奇数。将其分为两部分：1 和 arr[i]-1。由于 arr[i]-1 是偶数，因此可以通过上述策略使其为 0。因此，奇数元素可以将其大小减少到 1。因此，可以通过遵循上述策略使两个奇数元素为 0，最后将它们(即 1)异或(即 1)为 1 XOR 1 = 0。因此，如果数组中的奇数元素数是偶数，那么答案是可能的。否则，将留下值为 1 的元素，无法满足条件。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function that finds if it is
// possible to make the array
// contain only 1 element i.e. 0
string solve(vector& A)
{
    int i, ctr = 0;
    for (i = 0; i < A.size();
         i++) {
 
        // Check if element is odd
        if (A[i] % 2) {
            ctr++;
        }
    }
 
    // According to the logic
    // in above approach
    if (ctr % 2) {
        return "No";
    }
    else {
        return "Yes";
    }
}
 
// Driver code
int main()
{
 
    vector arr = { 9, 17 };
 
    cout << solve(arr) << endl;
    return 0;
}

Java

// Java program for the above approach
class GFG{
     
// Function that finds if it is
// possible to make the array
// contain only 1 element i.e. 0
public static String solve(int[] A)
{
    int i, ctr = 0;
         
    for(i = 0; i < A.length; i++)
    {
     
       // Check if element is odd
       if (A[i] % 2 == 1)
       {
           ctr++;
       }
    }
     
    // According to the logic
    // in above approach
    if (ctr % 2 == 1)
    {
        return "No";
    }
    else
    {
        return "Yes";
    }
}
 
// Driver code   
public static void main(String[] args)
{
    int[] arr = { 9, 17 };
    System.out.println(solve(arr));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
 
# Function that finds if it is
# possible to make the array
# contain only 1 element i.e. 0
def solve(A):
     
    ctr = 0
     
    for i in range(len(A)):
         
        # Check if element is odd
        if A[i] % 2 == 1:
            ctr += 1
             
    # According to the logic
    # in above approach
    if ctr % 2 == 1:
        return 'No'
    else :
        return 'Yes'
     
# Driver code
if __name__=='__main__':
     
    arr = [9, 17]
 
    print(solve(arr))
     
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function that finds if it is
// possible to make the array
// contain only 1 element i.e. 0
public static string solve(int[] A)
{
    int i, ctr = 0;
         
    for(i = 0; i < A.Length; i++)
    {
         
       // Check if element is odd
       if (A[i] % 2 == 1)
       {
           ctr++;
       }
    }
     
    // According to the logic
    // in above approach
    if (ctr % 2 == 1)
    {
        return "No";
    }
    else
    {
        return "Yes";
    }
}
 
// Driver code
public static void Main()
{
    int[] arr = { 9, 17 };
     
    Console.Write(solve(arr));
}
}
 
// This code is contributed by chitranayal

Javascript

 
// Javascript program for the above approach
 
// Function that finds if it is
// possible to make the array
// contain only 1 element i.e. 0
function solve(A)
{
    let i, ctr = 0;
    for (i = 0; i < A.length;
         i++) {
 
        // Check if element is odd
        if (A[i] % 2) {
            ctr++;
        }
    }
 
    // According to the logic
    // in above approach
    if (ctr % 2) {
        return "No";
    }
    else {
        return "Yes";
    }
}
 
// Driver code
 
    let arr = [ 9, 17 ];
 
    document.write(solve(arr));
 

输出：

Yes

时间复杂度： O(N)
辅助空间复杂度： O(1)

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