给定数组arr [] ，任务是计算其xor给出唯一素数的所有对，即没有两个对应该给出相同素数。
例子：
Input: arr[] = {2, 3, 4, 5, 6, 7, 8, 9}
Output: 6
(2, 5), (2, 7), (2, 9), (4, 6), (4, 7) and (4, 9) are the only pairs whose XORs are primes i.e. 7, 5, 11, 2, 3 and 13 respectively.
Input: arr[] = {10, 12, 23, 45, 5, 6}
Output: 4
方法：迭代每个可能的对，并检查当前对的xor是否为质数。如果它是素数，则更新count = count + 1并将素数保存在unordered_map中，以便跟踪重复的素数。最后打印计数。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function that returns true if n is prime
bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to return the count of valid pairs
int countPairs(int a[], int n)
{
int count = 0;
unordered_map m;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// If xor(a[i], a[j]) is prime and unique
if (isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == 0) {
m[(a[i] ^ a[j])]++;
count++;
}
}
}
return count;
}
// Driver code
int main()
{
int a[] = { 10, 12, 23, 45, 5, 6 };
int n = sizeof(a) / sizeof(a[0]);
cout << countPairs(a, n);
} Java
// Java implementation of above approach
import java.util.*;
class solution
{
// Function that returns true if n is prime
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to return the count of valid pairs
static int countPairs(int a[], int n)
{
int count = 0;
Map m=new HashMap< Integer,Integer>();
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// If xor(a[i], a[j]) is prime and unique
if (isPrime(a[i] ^ a[j]) && m.get(a[i] ^ a[j]) == null) {
m.put((a[i] ^ a[j]),1);
count++;
}
}
}
return count;
}
// Driver code
public static void main(String args[])
{
int a[] = { 10, 12, 23, 45, 5, 6 };
int n = a.length;
System.out.println(countPairs(a, n));
}
}
// This code is contributed by Arnab Kundu Python3
# Python3 implementation of above approach
# Function that returns true if n is prime
def isPrime(n):
# Corner cases
if n <= 1:
return False
if n <= 3:
return True
# This is checked so that we can skip
# middle five numbers in below loop
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
return True
# Function to return the count of valid pairs
def countPairs(a, n):
count = 0
m = dict()
for i in range(n - 1):
for j in range(i + 1, n):
# If xor(a[i], a[j]) is prime and unique
if isPrime(a[i] ^ a[j]) and m.get(a[i] ^ a[j], 0) == 0:
m[(a[i] ^ a[j])] = 1
count += 1
return count
# Driver code
a = [10, 12, 23, 45, 5, 6]
n = len(a)
print(countPairs(a, n))
# This code is contributed by
# Rajnis09C#
// C# implementation of the approach
using System ;
using System.Collections ;
class solution
{
// Function that returns true if n is prime
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to return the count of valid pairs
static int countPairs(int []a, int n)
{
int count = 0;
Hashtable m=new Hashtable();
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// If xor(a[i], a[j]) is prime and unique
if (isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == null)
{
m.Add((a[i] ^ a[j]),1);
count++;
}
}
}
return count;
}
// Driver code
public static void Main()
{
int []a = { 10, 12, 23, 45, 5, 6 };
int n = a.Length;
Console.WriteLine(countPairs(a, n));
}
// This code is contributed by Ryuga
}Javascript
输出：
4