GCD等于给定阵列的GCD的最小子序列

给定大小为N的数组arr [] ，任务是查找给定数组的最小子序列，该子序列的GCD等于给定数组的GCD 。如果存在多个这样的子序列，请打印其中的任何一个。

例子：

Input: arr[] = {4, 6, 12}
Output: 4 6
Explanation: Smallest subsequence having gcd equal to gcd of the given array(= 2) is {4, 6}. Therefore, the required output is {4, 6}

Input: arr[] = {6, 12, 18, 24}
Output: 6

天真的方法：解决此问题的最简单方法是生成给定数组的所有可能子序列，并计算每个子序列的GCD。打印gcd等于给定数组的gcd的最小子序列。

时间复杂度： O(2 ^N * N * log X)，其中X是给定数组的最大元素。
辅助空间： O(1)

高效方法：为了优化上述方法，该思想基于以下观察结果：

Length of the smallest subsequence having gcd equal to gcd of the given array must be either 1 or 2.

Consider, gcd of the given array is Y.

If arr[idx] = Y, then length of the smallest subsequence must be 1 for some value of idx.

Otherwise, length of the smallest subsequence must be 2.

Proof using Contradiction method:
If gcd of all possible pairs of the given array is greater than Y, then gcd of the given array must be greater than Y, which is not possible.
Therefore, at least one pair exists in the given array whose gcd is equal to Y.

请按照以下步骤解决问题：

• 初始化变量gcdArr以存储阵列的GCD。
• 遍历给定的数组，并检查是否有任何数组元素等于gcdArr 。如果发现为真，则打印该元素。
• 否则，从给定数组打印一对，其gcd等于gcdArr 。

下面是上述方法的实现。

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to print
// the smallest subsequence
// that satisfies the condition
void printSmallSub(int arr[], int N)
{
    // Stores gcd of the array.
    int gcdArr = 0;
 
    // Traverse the given array
    for (int i = 0; i < N; i++) {
 
        // Update gcdArr
        gcdArr = __gcd(gcdArr, arr[i]);
    }
 
    // Traverse the given array.
    for (int i = 0; i < N; i++) {
 
        // If current element
        // equal to gcd of array.
        if (arr[i] == gcdArr) {
            cout << arr[i] << " ";
            return;
        }
    }
 
    // Generate all possible pairs.
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N;
             j++) {
 
            // If gcd of current pair
            // equal to gcdArr
            if (__gcd(arr[i], arr[j])
                == gcdArr) {
 
                // Print current pair
                // of the array
                cout << arr[i] << " " << arr[j];
                return;
            }
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 6, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    printSmallSub(arr, N);
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to calculate gcd
// of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
         
    return gcd(b, a % b);
}
 
// Function to print
// the smallest subsequence
// that satisfies the condition
static void printSmallSub(int[] arr, int N)
{
     
    // Stores gcd of the array.
    int gcdArr = 0;
 
    // Traverse the given array
    for(int i = 0; i < N; i++)
    {
         
        // Update gcdArr
        gcdArr = gcd(gcdArr, arr[i]);
    }
 
    // Traverse the given array.
    for(int i = 0; i < N; i++)
    {
         
        // If current element
        // equal to gcd of array.
        if (arr[i] == gcdArr)
        {
            System.out.print(arr[i] + " ");
            return;
        }
    }
 
    // Generate all possible pairs.
    for(int i = 0; i < N; i++)
    {
        for(int j = i + 1; j < N; j++)
        {
             
            // If gcd of current pair
            // equal to gcdArr
            if (gcd(arr[i], arr[j]) == gcdArr)
            {
                 
                // Print current pair
                // of the array
                System.out.print(arr[i] + " " +
                                 arr[j]);
                return;
            }
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 4, 6, 12 };
    int N = arr.length;
 
    printSmallSub(arr, N);
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program to implement
# the above approach
import math
 
# Function to print the
# smallest subsequence
# that satisfies the condition
def printSmallSub(arr, N):
     
    # Stores gcd of the array.
    gcdArr = 0
 
    # Traverse the given array
    for i in range(0, N):
         
        # Update gcdArr
        gcdArr = math.gcd(gcdArr, arr[i])
 
    # Traverse the given array.
    for i in range(0, N):
         
        # If current element
        # equal to gcd of array.
        if (arr[i] == gcdArr):
            print(arr[i], end = " ")
            return
 
        # Generate all possible pairs.
        for i in range(0, N):
            for j in range(i + 1, N):
                 
                # If gcd of current pair
                # equal to gcdArr
                if (math.gcd(arr[i],
                             arr[j]) == gcdArr):
 
                    # Print current pair
                    # of the array
                    print(arr[i], end = " ")
                    print(arr[j], end = " ")
                    return
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 4, 6, 12 ]
    N = len(arr)
     
    printSmallSub(arr, N)
 
# This code is contributed by akhilsaini

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to calculate gcd
// of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
         
    return gcd(b, a % b);
}
 
// Function to print the
// smallest subsequence
// that satisfies the condition
static void printSmallSub(int[] arr, int N)
{
     
    // Stores gcd of the array.
    int gcdArr = 0;
 
    // Traverse the given array
    for(int i = 0; i < N; i++)
    {
         
        // Update gcdArr
        gcdArr = gcd(gcdArr, arr[i]);
    }
 
    // Traverse the given array.
    for(int i = 0; i < N; i++)
    {
         
        // If current element
        // equal to gcd of array.
        if (arr[i] == gcdArr)
        {
            Console.Write(arr[i] + " ");
            return;
        }
    }
 
    // Generate all possible pairs.
    for(int i = 0; i < N; i++)
    {
        for(int j = i + 1; j < N; j++)
        {
             
            // If gcd of current pair
            // equal to gcdArr
            if (gcd(arr[i], arr[j]) == gcdArr)
            {
                 
                // Print current pair
                // of the array
                Console.Write(arr[i] + " " +
                              arr[j]);
                return;
            }
        }
    }
}
 
// Driver Code
public static void Main()
{
 
    int[] arr = { 4, 6, 12 };
    int N = arr.Length;
 
    printSmallSub(arr, N);
}
}
 
// This code is contributed by akhilsaini

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to calculate gcd
// of two numbers
function gcd(a, b)
{
    if (b == 0)
        return a;
          
    return gcd(b, a % b);
}
 
// Function to print
// the smallest subsequence
// that satisfies the condition
function printSmallSub(arr, N)
{
    // Stores gcd of the array.
    let gcdArr = 0;
 
    // Traverse the given array
    for (let i = 0; i < N; i++) {
 
        // Update gcdArr
        gcdArr = gcd(gcdArr, arr[i]);
    }
 
    // Traverse the given array.
    for (let i = 0; i < N; i++) {
 
        // If current element
        // equal to gcd of array.
        if (arr[i] == gcdArr) {
            document.write(arr[i] + " ");
            return;
        }
    }
 
    // Generate all possible pairs.
    for (let i = 0; i < N; i++) {
        for (let j = i + 1; j < N;
            j++) {
 
            // If gcd of current pair
            // equal to gcdArr
            if (gcd(arr[i], arr[j])
                == gcdArr) {
 
                // Print current pair
                // of the array
                document.write(arr[i] + " " + arr[j]);
                return;
            }
        }
    }
}
 
// Driver Code
 
    let arr = [ 4, 6, 12 ];
    let N = arr.length;
 
    printSmallSub(arr, N);
 
// This is code is contributed by Mayank Tyagi
 

输出：

4 6

时间复杂度： (N ² * log X)，其中X是给定数组的最大元素。
辅助空间： O(1)