检查所有数组元素是否成对互质

给定一个由N 个正整数组成的数组A[] ，任务是检查所有数组元素是否都是成对互质的，即对于所有对 (A _i , A _j )，使得 1<=iGCD(A _i , A _j ) = 1 。

例子：

Input : A[] = {2, 3, 5}
Output : Yes
Explanation : All the pairs, (2, 3), (3, 5), (2, 5) are pairwise co-prime.

Input : A[] = {5, 10}
Output : No
Explanation : GCD(5, 10)=5 so they are not co-prime.

朴素的方法：解决问题的最简单的方法是从给定的数组中生成所有可能的对，对于每一对，检查它是否是互质的。如果发现任何对是非互质的，则打印“否”。否则，打印“是”。
时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：上述方法可以基于以下观察进行优化：

If any two numbers have a common prime factor, then their GCD can never be 1.  

这也可以解释为：

The LCM of the array must be equal to the product of the elements in the array.

因此，该解决方案归结为计算给定数组的 LCM 并检查它是否等于所有数组元素的乘积。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
#define ll long long int
 
// Function to calculate GCD
ll GCD(ll a, ll b)
{
    if (a == 0)
        return b;
    return GCD(b % a, a);
}
 
// Function to calculate LCM
ll LCM(ll a, ll b)
{
    return (a * b)
        / GCD(a, b);
}
 
// Function to check if all elements
// in the array are pairwise coprime
void checkPairwiseCoPrime(int A[], int n)
{
    // Initialize variables
    ll prod = 1;
    ll lcm = 1;
 
    // Iterate over the array
    for (int i = 0; i < n; i++) {
 
        // Calculate product of
        // array elements
        prod *= A[i];
 
        // Calculate LCM of
        // array elements
        lcm = LCM(A[i], lcm);
    }
 
    // If the product of array elements
    // is equal to LCM of the array
    if (prod == lcm)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
// Driver Code
int main()
{
    int A[] = { 2, 3, 5 };
    int n = sizeof(A) / sizeof(A[0]);
 
    // Function call
    checkPairwiseCoPrime(A, n);
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to calculate GCD
static long GCD(long a, long b)
{
    if (a == 0)
        return b;
         
    return GCD(b % a, a);
}
 
// Function to calculate LCM
static long LCM(long a, long b)
{
    return (a * b) / GCD(a, b);
}
 
// Function to check if all elements
// in the array are pairwise coprime
static void checkPairwiseCoPrime(int A[], int n)
{
     
    // Initialize variables
    long prod = 1;
    long lcm = 1;
 
    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
         
        // Calculate product of
        // array elements
        prod *= A[i];
 
        // Calculate LCM of
        // array elements
        lcm = LCM(A[i], lcm);
    }
     
    // If the product of array elements
    // is equal to LCM of the array
    if (prod == lcm)
        System.out.println("Yes");
    else
        System.out.println("No");
}
 
// Driver Code
public static void main (String[] args)
{
    int A[] = { 2, 3, 5 };
    int n = A.length;
     
    // Function call
    checkPairwiseCoPrime(A, n);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to calculate GCD
def GCD(a, b):
     
    if (a == 0):
        return b
         
    return GCD(b % a, a)
 
# Function to calculate LCM
def LCM(a, b):
     
    return (a * b) // GCD(a, b)
 
# Function to check if aelements
# in the array are pairwise coprime
def checkPairwiseCoPrime(A, n):
     
    # Initialize variables
    prod = 1
    lcm = 1
 
    # Iterate over the array
    for i in range(n):
 
        # Calculate product of
        # array elements
        prod *= A[i]
 
        # Calculate LCM of
        # array elements
        lcm = LCM(A[i], lcm)
 
    # If the product of array elements
    # is equal to LCM of the array
    if (prod == lcm):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 2, 3, 5 ]
    n = len(A)
 
    # Function call
    checkPairwiseCoPrime(A, n)
 
# This code is contributed by mohit kumar 29

C#

// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to calculate GCD
static long GCD(long a,
                long b)
{
  if (a == 0)
    return b;
  return GCD(b % a, a);
}
 
// Function to calculate LCM
static long LCM(long a,
                long b)
{
  return (a * b) / GCD(a, b);
}
 
// Function to check if all elements
// in the array are pairwise coprime
static void checkPairwiseCoPrime(int []A,
                                 int n)
{    
  // Initialize variables
  long prod = 1;
  long lcm = 1;
 
  // Iterate over the array
  for(int i = 0; i < n; i++)
  {
    // Calculate product of
    // array elements
    prod *= A[i];
 
    // Calculate LCM of
    // array elements
    lcm = LCM(A[i], lcm);
  }
 
  // If the product of array elements
  // is equal to LCM of the array
  if (prod == lcm)
    Console.WriteLine("Yes");
  else
    Console.WriteLine("No");
}
 
// Driver Code
public static void Main(String[] args)
{
  int []A = {2, 3, 5};
  int n = A.Length;
 
  // Function call
  checkPairwiseCoPrime(A, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// javascript program for the above approach   
// Function to calculate GCD
    function GCD(a , b) {
        if (a == 0)
            return b;
 
        return GCD(b % a, a);
    }
 
    // Function to calculate LCM
    function LCM(a , b) {
        return (a * b) / GCD(a, b);
    }
 
    // Function to check if all elements
    // in the array are pairwise coprime
    function checkPairwiseCoPrime(A , n) {
 
        // Initialize variables
        var prod = 1;
        var lcm = 1;
 
        // Iterate over the array
        for (i = 0; i < n; i++) {
 
            // Calculate product of
            // array elements
            prod *= A[i];
 
            // Calculate LCM of
            // array elements
            lcm = LCM(A[i], lcm);
        }
 
        // If the product of array elements
        // is equal to LCM of the array
        if (prod == lcm)
            document.write("Yes");
        else
            document.write("No");
    }
 
    // Driver Code
     
        var A = [ 2, 3, 5 ];
        var n = A.length;
 
        // Function call
        checkPairwiseCoPrime(A, n);
 
// This code contributed by umadevi9616

输出：

Yes

时间复杂度： O(N log (min(A[i])))
辅助空间： O(1)

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