通过对给定 Array 重新排序，GCD 大于 1 的 Array 中的最大对数

给定一个大小为N的数组arr[] 。任务是重新排序arr[]并找到遵循以下条件的最大 GCD 对数。

• 选择数组Ai 和 Aj的任意两个元素，其中 0 <= i < j < N。
• 将Aj 与大于 1的(Ai, 2 * Aj)相乘后计算GCD 。

例子

Input: arr[] = { 3, 6 . 5, 3}
Output: 4
Explanation: Reorder array like this : { 6, 5, 3, 3 } and below are the pairs formed from arr[].
P1 = GCD( 6, 5 * 2) => (6, 10) => 2 > 1
P2 = GCD( 6, 3 * 2) => (6, 6) => 6 > 1
P3 = GCD( 6, 3 * 2) => (6, 6) => 6 > 1
P4 = GCD( 3, 3 * 2) => (3, 6) => 3 > 1

Input: arr[] = { 1, 7 }
Output: 0
Explanation: If array is order like { 7, 1 } no pair can be formed
GCD(7, 1 * 2) = > (7, 2 ), GCD(1, 7 * 2) => (1, 14) == 1

方法：如果我们观察到如果偶数元素在起始位置，那么(GCD > 1)对是最大的，因为存在乘以arr[j] * 2的条件，并且奇数元素的顺序与其数量无关对总是相同的。请按照以下步骤解决给定的问题。

• 用值0初始化变量idx 。
• 遍历数组。
• 如果arr[i]是偶数，则与arr[idx]交换并将 idx 增加1 。
• 遍历所有元素后。
• 用0初始化变量ans 。
• 首先使用两个循环i和第二个使用j = i + 1 。
• 现在如果gcd(arr[i], 2 * arr[j] * 2) > 1将ans增加1 。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find the maximum number of pairs
int maximumpairs(int arr[], int n)
{
 
    // Reorder array with even element first
    int idx = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) {
            swap(arr[i], arr[idx]);
            idx++;
        }
    }
 
    // Now count the ans
    int ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (__gcd(arr[i], 2 * arr[j]) > 1) {
                ans++;
            }
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
 
    // Initializations
    int arr[] = { 5, 3, 6, 3 };
    int N = sizeof(arr) / sizeof(int);
 
    // Function Call
    int ans = maximumpairs(arr, N);
 
    cout << ans;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
  // find gcd
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
   
  // Function to find the maximum number of pairs
  static int maximumpairs(int arr[], int n)
  {
 
    // Reorder array with even element first
    int idx = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] % 2 == 0) {
        //swap
        int temp = arr[i];
        arr[i] = arr[idx];
        arr[idx] = temp;
        idx++;
      }
    }
 
    // Now count the ans
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int a = arr[i];
        int b = 2 * arr[j];
        if (gcd(a, b) > 1) {
          ans++;
        }
      }
    }
    return ans;
  }
 
  public static void main (String[] args)
  {
 
    // Initializations
    int arr[] = { 5, 3, 6, 3 };
    int N = arr.length;
 
    // Function Call
    int ans = maximumpairs(arr, N);
 
    System.out.println(ans);
  }
}
 
// This code is contributed by hrithikgarg03188

Python3

# Python program for above approach
 
# Function to find GCD
def gcd(a, b):
    if(b == 0):
        return a
    else:
        return gcd(b, a % b)
         
# Function to find the maximum number of pairs
def maximumpairs(arr,n):
 
    # Reorder array with even element first
    idx = 0
    for i in range(0, n):
        if (arr[i] % 2 == 0):
            arr[i], arr[idx] = arr[idx], arr[i]
            idx = idx + 1
 
    # Now count the ans
    ans = 0
    for i in range(0,n):
        for j in range(i + 1,n):
            if (gcd(arr[i], 2*arr[j]) > 1):
                ans = ans + 1
             
    return ans
 
# Driver Code
 
# Initializations
arr = [ 5, 3, 6, 3 ]
N = len(arr)
 
# Function Call
ans = maximumpairs(arr, N)
 
print(ans)
     
# This code is contributed by Taranpreet

C#

// C# program for the above approach
using System;
 
class GFG {
  // find gcd
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
   
  // Function to find the maximum number of pairs
  static int maximumpairs(int []arr, int n)
  {
 
    // Reorder array with even element first
    int idx = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] % 2 == 0) {
        //swap
        int temp = arr[i];
        arr[i] = arr[idx];
        arr[idx] = temp;
        idx++;
      }
    }
 
    // Now count the ans
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int a = arr[i];
        int b = 2 * arr[j];
        if (gcd(a, b) > 1) {
          ans++;
        }
      }
    }
    return ans;
  }
 
  public static void Main ()
  {
 
    // Initializations
    int []arr = { 5, 3, 6, 3 };
    int N = arr.Length;
 
    // Function Call
    int ans = maximumpairs(arr, N);
 
    Console.Write(ans);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// Javascript program for the above approach
 
// find gcd
function gcd(a, b)
{
  if (b == 0)
    return a;
  return gcd(b, a % b);
}
 
// Function to find the maximum number of pairs
function maximumpairs(arr, n)
{
 
  // Reorder array with even element first
  let idx = 0;
  for (let i = 0; i < n; i++) {
    if (arr[i] % 2 == 0) {
      //swap
      let temp = arr[i];
      arr[i] = arr[idx];
      arr[idx] = temp;
      idx++;
    }
  }
 
  // Now count the ans
  let ans = 0;
  for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {
      let a = arr[i];
      let b = 2 * arr[j];
      if (gcd(a, b) > 1) {
        ans++;
      }
    }
  }
  return ans;
}
 
  // Initializations
  let arr = [ 5, 3, 6, 3 ];
  let N = arr.length;
 
  // Function Call
  let ans = maximumpairs(arr, N);
 
  document.write(ans);
 
 // This code is contributed by Samim Hossain Mondal.
 

输出

4

时间复杂度： O(N * N * logN)
辅助空间： O(1)