查找数组中所有对的最小 GCD

给定一个正整数数组arr ，任务是为给定数组的任何对找到可能的最小 GCD。
例子：

Input: arr[] = {1, 2, 3, 4, 5} 
Output: 1 
Explanation: 
GCD(1, 2) = 1.
Input: arr[] = {2, 4, 6, 8, 3} 
Output: 1 

方法：
如果我们观察清楚，我们会注意到任意两个数的最小 GCD 将是数组中所有元素的 GCD。这种观察背后的原因是，如果使整个阵列的 GCD 最小的元素出现在该对中，则任何对的 GCD 最小。

Illustration: 
Let us consider an array arr[] = {2,4,6,8,3}. 
GCD of the all array elements excluding 3 is 2. On considering 3, the gcd minimizes to 1. 
GCD of all the possible pairs are: 
gcd(2,4) = 2 
gcd(2,6) = 2 
gcd(2,8) = 2 
gcd(2,3) = 1 
gcd(4,6) = 2 
gcd(4,8) = 4 
gcd(4,3) = 1 
gcd(6,3) = 3 
gcd(6,8) = 2 
gcd(8,3) = 1 
Thus, the minimum gcd for any pair is equal to 1, which is equal to gcd of the array and only appears for the pairs which contain 3 in it. 

下面的代码是上述方法的实现：

C++

// C++ program to find the
// minimum GCD of any pair
// in the array
#include 
using namespace std;
 
// Function returns the
// Minimum GCD of any pair
int MinimumGCD(int arr[], int n)
{
    int g = 0;
 
    // Finding GCD of all the
    // elements in the array.
    for (int i = 0; i < n; i++) {
        g = __gcd(g, arr[i]);
    }
    return g;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4, 6, 8, 3 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << MinimumGCD(arr, N) << endl;
}

Java

// Java program to find the
// minimum GCD of any pair
// in the array
import java.util.*;
 
class GFG{
     
static int __gcd(int a, int b)
{
    if(b == 0)
        return a;
    else
        return __gcd(b, a % b);
}
 
// Function returns the
// minimum GCD of any pair
static int MinimumGCD(int arr[], int n)
{
    int g = 0;
 
    // Finding GCD of all the
    // elements in the array.
    for(int i = 0; i < n; i++)
    {
       g = __gcd(g, arr[i]);
    }
     
    return g;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 6, 8, 3 };
    int N = arr.length;
 
    System.out.println(MinimumGCD(arr, N));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the
# minimum GCD of any pair
# in the array
 
from math import gcd
 
# Function returns the
# Minimum GCD of any pair
def MinimumGCD(arr, n):
    g = 0
 
    # Finding GCD of all the
    # elements in the array.
    for i in range(n):
        g = gcd(g, arr[i])
    return g
 
# Driver code
if __name__ == '__main__':
     
    arr = [ 2, 4, 6, 8, 3 ]
 
    N = len(arr)
    print(MinimumGCD(arr, N))
 
# This code is contributed by Samarth

C#

// C# program to find the
// minimum GCD of any pair
// in the array
using System;
 
class GFG{
     
static int __gcd(int a, int b)
{
    if(b == 0)
        return a;
    else
        return __gcd(b, a % b);
}
 
// Function returns the
// minimum GCD of any pair
static int MinimumGCD(int []arr, int n)
{
    int g = 0;
 
    // Finding GCD of all the
    // elements in the array.
    for(int i = 0; i < n; i++)
    {
       g = __gcd(g, arr[i]);
    }
     
    return g;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 4, 6, 8, 3 };
    int N = arr.Length;
 
    Console.WriteLine(MinimumGCD(arr, N));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// JavaScript program to find the
// minimum GCD of any pair
// in the array
 
function calGCD(a, b) {
  if (!b) {
    return a;
  }
 
  return calGCD(b, a % b);
}
 
// Function returns the
// Minimum GCD of any pair
function MinimumGCD(arr, n)
{
    var g = 0;
    var i;
    // Finding GCD of all the
    // elements in the array.
    for (i = 0; i < n; i++) {
        g = calGCD(g, arr[i]);
    }
    return g;
}
 
// Driver code
    var arr = [2, 4, 6, 8, 3];
    var N = arr.length;
    document.write(MinimumGCD(arr, N));
 

输出：

1

时间复杂度： O(N)
辅助空间： O(1)

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