从给定数组中找到 N – 1 对，使得所有对和的 GCD 大于 1

给定一个由2 * N 个整数组成的数组arr[] ，任务是找到一组N-1对，使得所有对和的 GCD 都大于 1。

例子：

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output:
1 3
2 4
Explanation: The given array has 3 * 2 elements. The pair (1, 3) and (2, 4) has sum of elements as 4 and 6 respectively. Also, GCD(4, 6) > 1. Hence, (1, 3) and (2, 4) are the required pairs. Other possible pairs are (1, 5) and (2, 6), (3, 5) and (2, 5), etc..

Input: arr[] = {1, 1, 1, 2, 3, 4}
Output:
1 3
2 4

编程需要懂一点英语

方法：给定的问题是一个基于观察的问题。这个想法是创建N - 1对，使得它们的总和是偶数。此外，对于具有偶数和的一对，要么两个整数都应该是偶数，要么都应该是奇数。以下是要遵循的步骤：

初始化两个向量Odd和Even分别存储奇数和偶数。
遍历给定数组arr[]并在奇数向量中插入奇数整数，在偶数向量中插入偶数整数。
将向量奇数和偶数的连续元素配对并打印N – 1第一对整数。

下面是上述方法的实现：

C++

// C++ Program of the above approach
#include 
using namespace std;
 
// Function to find N - 1 pairs from
// given array such that GCD of all
// pair-sums is greater than 1
void findPairs(int arr[], int N)
{
    // Stores even and odd
    // integers respectively
    vector even, odd;
 
    // Loop to iterate arr[]
    for (int i = 0; i < 2 * N; i++) {
        if (arr[i] & 1)
            odd.push_back(arr[i]);
        else
            even.push_back(arr[i]);
    }
 
    // Stores the required pairs
    vector > ans;
 
    // Insert all possible pairs
    // from odd elements
    for (int i = 0; i + 1 < odd.size(); i += 2)
        ans.push_back({ odd[i], odd[i + 1] });
 
    // Insert all possible pairs
    // from even elements
    for (int i = 0; i + 1 < even.size(); i += 2)
        ans.push_back({ even[i], even[i + 1] });
 
    // Print Answer
    for (int i = 0; i < N - 1; i++) {
        cout << ans[i].first << " " << ans[i].second
             << endl;
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = 3;
    findPairs(arr, N);
 
    return 0;
}

Java

// JAVA Program of the above approach
import java.util.*;
class Pair {
  int x;
  int y;
 
  // Constructor
  public Pair(int x, int y)
  {
    this.x = x;
    this.y = y;
  }
}
class GFG
{
 
  // Function to find N - 1 pairs from
  // given array such that GCD of all
  // pair-sums is greater than 1
  public static void findPairs(int arr[], int N)
  {
 
    // Stores even and odd
    // integers respectively
    ArrayList even = new ArrayList();
    ArrayList odd = new ArrayList();
 
    // Loop to iterate arr[]
    for (int i = 0; i < 2 * N; i++) {
      if (arr[i] % 2 == 1)
        odd.add(arr[i]);
      else
        even.add(arr[i]);
    }
 
    // Stores the required pairs
    Pair[] ans1 = new Pair[N];
    Pair[] ans2 = new Pair[N];
 
    // Insert all possible pairs
    // from odd elements
    for (int i = 0; i + 1 < odd.size(); i += 2) {
      ans1[i] = new Pair(odd.get(i), odd.get(i + 1));
    }
 
    // Insert all possible pairs
    // from even elements
    for (int i = 0; i + 1 < even.size(); i += 2) {
      ans2[i]
        = new Pair(even.get(i), even.get(i + 1));
    }
 
    // Print Answer
    for (int i = 0; i < N - 1; i++) {
      System.out.println(ans1[i].x + " " + ans1[i].y);
      System.out.println(ans2[i].x + " " + ans2[i].y);
      break;
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };
    int N = 3;
    findPairs(arr, N);
  }
}
 
// This code is contributed by Taranpreet

Python3

# Python code for the above approach
 
# Function to find N - 1 pairs from
# given array such that GCD of all
# pair-sums is greater than 1
def findPairs(arr, N):
 
    # Stores even and odd
    # integers respectively
    even = []
    odd = [];
 
    # Loop to iterate arr[]
    for i in range(2 * N):
        if (arr[i] & 1):
            odd.append(arr[i]);
        else:
            even.append(arr[i]);
     
 
    # Stores the required pairs
    ans = [];
 
    # Insert all possible pairs
    # from odd elements
    i = 0;
    while(i + 1 < len(odd)):
        ans.append({ "first": odd[i], "second": odd[i + 1] });
        i += 2
 
    # Insert all possible pairs
    # from even elements
    i = 0
    while(i + 1 < len(odd)):
        ans.append({ "first": even[i], "second": even[i + 1] });
        i += 2
 
    # Print Answer
    for i in range(N - 1):
        for y in ans[i].values():
            print(y, end=" ")
        print("")
 
# Driver Code
arr = [1, 2, 3, 4, 5, 6];
N = 3;
findPairs(arr, N);
 
# This code is contributed by Saurabh Jaiswal

C#

// C# Program of the above approach
using System;
using System.Collections.Generic;
 
public class Pair {
  public int x;
  public int y;
 
  // Constructor
  public Pair(int x, int y) {
    this.x = x;
    this.y = y;
  }
}
 
public class GFG
{
 
  // Function to find N - 1 pairs from
  // given array such that GCD of all
  // pair-sums is greater than 1
  public static void findPairs(int []arr, int N) {
 
    // Stores even and odd
    // integers respectively
    List even = new List();
    List odd = new List();
 
    // Loop to iterate []arr
    for (int i = 0; i < 2 * N; i++) {
      if (arr[i] % 2 == 1)
        odd.Add(arr[i]);
      else
        even.Add(arr[i]);
    }
 
    // Stores the required pairs
    Pair[] ans1 = new Pair[N];
    Pair[] ans2 = new Pair[N];
 
    // Insert all possible pairs
    // from odd elements
    for (int i = 0; i + 1 < odd.Count; i += 2) {
      ans1[i] = new Pair(odd[i], odd[i + 1]);
    }
 
    // Insert all possible pairs
    // from even elements
    for (int i = 0; i + 1 < even.Count; i += 2) {
      ans2[i] = new Pair(even[i], even[i + 1]);
    }
 
    // Print Answer
    for (int i = 0; i < N - 1; i++) {
      Console.WriteLine(ans1[i].x + " " + ans1[i].y);
      Console.WriteLine(ans2[i].x + " " + ans2[i].y);
      break;
    }
  }
 
  // Driver Code
  public static void Main(String[] args) {
    int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };
    int N = 3;
    findPairs(arr, N);
  }
}
 
// This code is contributed by Rajput-Ji

Javascript

输出

1 3
2 4

时间复杂度： O(N)
辅助空间： O(N)