给定一个由N 个正整数组成的数组arr[] ，任务是找到对的数量，使得对的最大公约数 (GCD) 不是素数。对(i, j)和(j, i)被认为是相同的。
例子：
Input: arr[] ={ 2, 3, 9}
Output: 10
Explanation:
Following are the possible pairs whose GCD is not prime:
- (0, 1): The GCD of arr[0](= 2) and arr[1](= 3) is 1.
- (0, 2): The GCD of arr[0](= 2) and arr[2](= 9) is 1.
Therefore, the total count of pairs is 2.
Input: arr[] = {3, 5, 2, 10}
Output: 4
方法：给定的问题可以通过找到直到10 5 的所有素数并将它们存储在一个集合中来解决给定的问题，然后对于每对(i, j)如果它们的 GCD 不在集合中，则计算这对。请按照以下步骤解决问题：
- 定义一个函数primeSieve()并使用 Eratosthenes 筛法查找最多10 5 的素数并将其存储在一个无序集合中，例如S 。
- 初始化变量，比如count为0 ，它存储对的总计数。
- 从给定数组生成所有可能的对，如果它们的 GCD 不在集合中，则将count的值增加1 。
- 执行上述步骤后，打印count的值作为答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the prime numbers
void primeSieve(bool* p)
{
for (int i = 2; i * i <= 1000000; i++) {
// If p[i] is not changed,
// then it is a prime
if (p[i] == true) {
// Update all multiples of i
// as non prime
for (int j = i * 2;
j <= 1000000; j += i) {
p[j] = false;
}
}
}
}
// Function to find GCD of two integers
// a and b
int gcd(int a, int b)
{
// Base Case
if (b == 0)
return a;
// Find GCD Recursively
return gcd(b, a % b);
}
// Function to count the number of
// pairs whose GCD is non prime
int countPairs(int arr[], int n,
unordered_set s)
{
// Stores the count of valid pairs
int count = 0;
// Traverse over the array arr[]
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Calculate the GCD
int x = gcd(arr[i], arr[j]);
// Update the count
if (s.find(x) == s.end())
count++;
}
}
// Return count
return count;
}
// Utility Function to find all the prime
// numbers and find all the pairs
void countPairsUtil(int arr[], int n)
{
// Stores all the prime numbers
unordered_set s;
bool p[1000005];
memset(p, true, sizeof(p));
// Find all the prime numbers
primeSieve(p);
s.insert(2);
// Insert prime numbers in the
// unordered set
for (int i = 3; i <= 1000000; i += 2)
if (p[i])
s.insert(i);
// Find the count of valid pairs
cout << countPairs(arr, n, s);
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 9 };
int N = sizeof(arr) / sizeof(arr[0]);
countPairsUtil(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the prime numbers
static void primeSieve(boolean[] p)
{
for (int i = 2; i * i <= 1000000; i++) {
// If p[i] is not changed,
// then it is a prime
if (p[i] == true) {
// Update all multiples of i
// as non prime
for (int j = i * 2;
j <= 1000000; j += i) {
p[j] = false;
}
}
}
}
// Function to find GCD of two integers
// a and b
static int gcd(int a, int b)
{
// Base Case
if (b == 0)
return a;
// Find GCD Recursively
return gcd(b, a % b);
}
// Function to count the number of
// pairs whose GCD is non prime
static int countPairs(int arr[], int n,
HashSet s)
{
// Stores the count of valid pairs
int count = 0;
// Traverse over the array arr[]
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Calculate the GCD
int x = gcd(arr[i], arr[j]);
// Update the count
if (!s.contains(x))
count++;
}
}
// Return count
return count;
}
// Utility Function to find all the prime
// numbers and find all the pairs
static void countPairsUtil(int arr[], int n)
{
// Stores all the prime numbers
HashSet s = new HashSet();
boolean []p = new boolean[1000005];
for(int i=0;i C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Function to find the prime numbers
static void primeSieve(bool[] p)
{
for (int i = 2; i * i <= 1000000; i++) {
// If p[i] is not changed,
// then it is a prime
if (p[i] == true) {
// Update all multiples of i
// as non prime
for (int j = i * 2;
j <= 1000000; j += i) {
p[j] = false;
}
}
}
}
// Function to find GCD of two integers
// a and b
static int gcd(int a, int b)
{
// Base Case
if (b == 0)
return a;
// Find GCD Recursively
return gcd(b, a % b);
}
// Function to count the number of
// pairs whose GCD is non prime
static int countPairs(int []arr, int n,
HashSet s)
{
// Stores the count of valid pairs
int count = 0;
// Traverse over the array []arr
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Calculate the GCD
int x = gcd(arr[i], arr[j]);
// Update the count
if (!s.Contains(x))
count++;
}
}
// Return count
return count;
}
// Utility Function to find all the prime
// numbers and find all the pairs
static void countPairsUtil(int []arr, int n)
{
// Stores all the prime numbers
HashSet s = new HashSet();
bool []p = new bool[1000005];
for(int i = 0; i < p.Length; i++)
p[i] = true;
// Find all the prime numbers
primeSieve(p);
s.Add(2);
// Insert prime numbers in the
// unordered set
for (int i = 3; i <= 1000000; i += 2)
if (p[i])
s.Add(i);
// Find the count of valid pairs
Console.Write(countPairs(arr, n, s));
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 3, 9 };
int N = arr.Length;
countPairsUtil(arr, N);
}
}
// This code is contributed by 29AjayKumar 输出：
2时间复杂度： O((N 2 )*log N)
辅助空间： O(N)
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