给定大小为N的数组arr [] ，任务是找到给定数组中所有数字的不同素数。
例子：
Input: N = 3, arr[] = {12, 15, 18}
Output: 2 3 5
Exlpanation:
12 = 2 x 2 x 3
15 = 3 x 5
18 = 2 x 3 x 3
Distinct prime factors among the given numbers are 2, 3, 5.
Input: N = 9, arr[] = {2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 2 3 5 7
天真的方法：解决此问题的一种简单方法是找到数组中每个数字的素因。然后在这些素数因子之间找到不同的素数。
时间复杂度： O(N 2 )
高效的方法：一种有效的方法是首先使用Eratosthenes筛子找到所有达到给定极限的质数，然后将它们存储在数组中。对于质数数组中的每个质数，请检查输入数组中的任何数是否可整。如果可以将其整除，则将该素数存储在答案数组中。最后，对给定输入数组中的所有数字重复此过程后，返回答案数组。
下面是上述方法的实现：
CPP
#include
using namespace std;
//cppimplementation of the above approach
//Function to return an array
//of prime numbers upto n
//using Sieve of Eratosthenes
vector seive(int n){
vector prime (n + 1,0);
int p = 2;
while(p * p<= n){
if(prime[p]== 0){
for (int i=2*p;i allPrimes;
for (int i =2;i distPrime(vector arr, vector allPrimes){
//Creating an empty array
//to store distinct prime factors
vector list1;
//Iterating through all the
//prime numbers and check if
//any of the prime numbers is a
//factor of the given input array
for (int i : allPrimes){
for (int j :arr){
if(j % i == 0){
list1.push_back(i);
break;
}
}
}
return list1;
}
//Driver code
int main()
{
//Finding prime numbers upto 10000
//using Sieve of Eratosthenes
vector allPrimes = seive(10000);
vector arr = {15, 30, 60};
vector ans = distPrime(arr, allPrimes);
cout<<"[";
for(int i:ans) cout< Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
// Function to return an array
// of prime numbers upto n
// using Sieve of Eratosthenes
static ArrayList seive(int n){
ArrayList prime = new ArrayList();
for(int i = 0; i < n + 1; i++)
prime.add(0);
int p = 2;
while(p * p <= n){
if(prime.get(p) == 0){
for (int i = 2 * p; i < n + 1; i += p)
prime.set(i, 1);
}
p += 1;
}
ArrayList allPrimes = new ArrayList();
for (int i = 2; i < n; i++){
if (prime.get(i) == 0)
allPrimes.add(i);
}
return allPrimes;
}
// Function to return distinct
// prime factors from the given array
static ArrayList distPrime(ArrayList arr,
ArrayList allPrimes){
// Creating an empty array
// to store distinct prime factors
ArrayList list1 = new ArrayList();
// Iterating through all the
// prime numbers and check if
// any of the prime numbers is a
// factor of the given input array
for (int i = 0; i < allPrimes.size(); i++){
for (int j = 0; j < arr.size(); j++){
if(arr.get(j) % allPrimes.get(i) == 0){
list1.add(allPrimes.get(i));
break;
}
}
}
return list1;
}
// Driver code
public static void main(String args[])
{
// Finding prime numbers upto 10000
// using Sieve of Eratosthenes
ArrayList allPrimes = new ArrayList(seive(10000));
ArrayList arr = new ArrayList();
arr.add(15);
arr.add(30);
arr.add(60);
ArrayList ans = new ArrayList(distPrime(arr, allPrimes));
System.out.print("[");
for(int i = 0; i < ans.size(); i++)
System.out.print(ans.get(i) + " ");
System.out.print("]");
}
}
// This code is contributed by Surendra_Gangwar Python3
# Python3 implementation of the above approach
# Function to return an array
# of prime numbers upto n
# using Sieve of Eratosthenes
def seive(n):
prime =[True]*(n + 1)
p = 2
while(p * p<= n):
if(prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
allPrimes = [x for x in range(2, n)if prime[x]]
return allPrimes
# Function to return distinct
# prime factors from the given array
def distPrime(arr, allPrimes):
# Creating an empty array
# to store distinct prime factors
list1 = list()
# Iterating through all the
# prime numbers and check if
# any of the prime numbers is a
# factor of the given input array
for i in allPrimes:
for j in arr:
if(j % i == 0):
list1.append(i)
break
return list1
# Driver code
if __name__=="__main__":
# Finding prime numbers upto 10000
# using Sieve of Eratosthenes
allPrimes = seive(10000)
arr = [15, 30, 60]
ans = distPrime(arr, allPrimes)
print(ans)
# This code is contributed by mohit kumar 29C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return an array
// of prime numbers upto n
// using Sieve of Eratosthenes
static List seive(int n)
{
List prime = new List();
for(int i = 0; i < n + 1; i++)
prime.Add(0);
int p = 2;
while(p * p <= n)
{
if(prime[p] == 0)
{
for (int i = 2 * p; i < n + 1; i += p)
prime[i]= 1;
}
p += 1;
}
List allPrimes = new List();
for (int i = 2; i < n; i++){
if (prime[i] == 0)
allPrimes.Add(i);
}
return allPrimes;
}
// Function to return distinct
// prime factors from the given array
static List distPrime(List arr,
List allPrimes){
// Creating an empty array
// to store distinct prime factors
List list1 = new List();
// Iterating through all the
// prime numbers and check if
// any of the prime numbers is a
// factor of the given input array
for (int i = 0; i < allPrimes.Count; i++){
for (int j = 0; j < arr.Count; j++){
if(arr[j] % allPrimes[i] == 0){
list1.Add(allPrimes[i]);
break;
}
}
}
return list1;
}
// Driver code
public static void Main(string []args)
{
// Finding prime numbers upto 10000
// using Sieve of Eratosthenes
List allPrimes = new List(seive(10000));
List arr = new List();
arr.Add(15);
arr.Add(30);
arr.Add(60);
List ans = new List(distPrime(arr, allPrimes));
Console.Write("[");
for(int i = 0; i < ans.Count; i++)
Console.Write(ans[i] + " ");
Console.Write("]");
}
}
// This code is contributed by chitranayal 输出：
[2, 3, 5]