通过重复从其最大值中减去对的绝对差来使所有数组元素相等

给定一个由N个整数组成的数组arr [] ，任务是通过从数组中选择任意一对整数并将整数对中的较大整数替换为它们的绝对差数来使所有数组元素相等。打印所有数组元素的最终值。

例子：

Input: arr[] ={2, 3, 4}
Output: 1
Explanation: 
Step 1: Performing on the pair (2, 3) modifies arr[] = {2, 1, 4}
Step 2: Performing on the pair (2, 4) modifies arr[] = {2, 1, 2}
Step 3: Performing on the pair (2, 1) modifies {1, 1, 2}
Step 4: Performing on the pair (1, 2) modifies arr[] = {1, 1, 1}

Input: arr[] = {24, 60}
Output: 12

方法：从上面的问题陈述中可以看出，对于任何一对(a，b) ，绝对差都从最大元素中减去。然后，此操作类似于查找该对中的GCD。因此，从该观察结果可以明显看出，所有阵列元素都需要还原为阵列的GCD。请按照以下步骤解决问题：

• 将变量gcd初始化为1 。
• 遍历给定的数组，并遍历更新gcd时为：

 gcd = gcd(arr[i], gcd), where 0 ≤ i < N

• 在上述步骤之后， gcd的值是在将给定操作应用于每对不同的元素对之后所需的数组元素。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to return
// gcd of a and b
int gcd(int a, int b)
{
  // Base Case
  if (a == 0)
    return b;
 
  // Recursive Call
  return gcd(b % a, a);
}
 
// Function to find gcd of array
int findGCD(int arr[], int N)
{
  // Initialise the result
  int result = 0;
 
  // Traverse the array arr[]
  for (int i = 0; i < N; i++)
  {
    // Update result as gcd of
    // the result and arr[i]
    result = gcd(result, arr[i]);
 
    if (result == 1)
    {
      return 1;
    }
  }
 
  // Return the resultant GCD
    return result;
}
 
// Driver Code
int main()
{
  // Given array arr[]
  int arr[] = {2, 3, 4};
 
  int N = sizeof(arr) /
          sizeof(arr[0]);
 
  // Function Call
  cout << findGCD(arr, N);
  return 0;
}
 
// This code is contributed by 29AjayKumar

Java

// Java program for the above approach
 
public class GCD {
 
    // Function to return gcd of a and b
    static int gcd(int a, int b)
    {
        // Base Case
        if (a == 0)
            return b;
 
        // Recursive Call
        return gcd(b % a, a);
    }
 
    // Function to find gcd of array
    static int findGCD(int arr[], int N)
    {
        // Initialise the result
        int result = 0;
 
        // Traverse the array arr[]
        for (int element : arr) {
 
            // Update result as gcd of
            // the result and arr[i]
            result = gcd(result, element);
 
            if (result == 1) {
                return 1;
            }
        }
 
        // Return the resultant GCD
        return result;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array arr[]
        int arr[] = { 2, 3, 4 };
 
        int N = arr.length;
 
        // Function Call
        System.out.println(findGCD(arr, N));
    }
}

Python3

# Python3 program for the above approach
 
# Function to return gcd of a and b
def gcd(a, b):
     
    # Base Case
    if (a == 0):
        return b
 
    # Recursive call
    return gcd(b % a, a)
 
# Function to find gcd of array
def findGCD(arr, N):
     
    # Initialise the result
    result = 0
 
    # Traverse the array arr[]
    for element in arr:
 
        # Update result as gcd of
        # the result and arr[i]
        result = gcd(result, element)
 
        if (result == 1):
            return 1
 
    # Return the resultant GCD
    return result
 
# Driver Code
 
# Given array arr[]
arr = [ 2, 3, 4 ]
 
N = len(arr)
 
# Function call
print(findGCD(arr, N))
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to return gcd of a and b
static int gcd(int a, int b)
{
     
    // Base Case
    if (a == 0)
        return b;
 
    // Recursive call
    return gcd(b % a, a);
}
 
// Function to find gcd of array
static int findGCD(int[] arr, int N)
{
     
    // Initialise the result
    int result = 0;
 
    // Traverse the array arr[]
    foreach(int element in arr)
    {
 
        // Update result as gcd of
        // the result and arr[i]
        result = gcd(result, element);
 
        if (result == 1)
        {
            return 1;
        }
    }
 
    // Return the resultant GCD
    return result;
}
 
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int[] arr = { 2, 3, 4 };
 
    int N = arr.Length;
 
    // Function call
    Console.WriteLine(findGCD(arr, N));
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// JavaScript program for
// the above approach
 
    // Function to return gcd of a and b
    function gcd(a, b)
    {
        // Base Case
        if (a == 0)
            return b;
  
        // Recursive Call
        return gcd(b % a, a);
    }
  
    // Function to find gcd of array
    function findGCD(arr, N)
    {
        // Initialise the result
        let result = 0;
  
        // Traverse the array arr[]
        for (let element in arr) {
  
            // Update result as gcd of
            // the result and arr[i]
            result = gcd(result, element);
  
            if (result == 1) {
                return 1;
            }
        }
  
        // Return the resultant GCD
        return result;
    }
 
// Driver code
 
          // Given array arr[]
        let arr = [ 2, 3, 4 ];
  
        let N = arr.length;
  
        // Function Call
        document.write(findGCD(arr, N));
                             

输出

1

时间复杂度： O(N * logN)，其中N是给定数组的大小。
辅助空间： O(N)