给定N个正整数的数组arr [] ，任务是计算非素数与素数的XOR之间的绝对差。请注意， 1既不是素数也不是复合数。
例子：
Input: arr[] = {1, 3, 5, 10, 15, 7}
Output: 4
Xor of non-primes = 10 ^ 15 = 5
Xor of primes = 3 ^ 5 ^ 7 = 1
|5 – 1| = 4
Input: arr[] = {3, 4, 6, 7}
Output: 2
天真的方法：一个简单的解决方案是遍历数组并检查每个元素是否为素数。如果number是质数，则将其与X1表示质数的XOR，否则将XOR与X2表示非质数的XOR 。遍历整个数组后，取X1和X2之间的绝对差。
高效的方法：使用Eratosthenes筛子生成所有素数，直到数组的最大元素。现在，遍历数组并检查当前元素是否为质数。如果元素是素数，则将其与X1进行XOR运算，否则将其与X2进行XOR运算。最终打印abs(X1-X2) 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find the absolute difference
// between the XOR of non-primes and the
// XOR of primes in the given array
int calculateDifference(int arr[], int n)
{
// Find maximum value in the array
int max_val = *max_element(arr, arr + n);
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
vector prime(max_val + 1, true);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Store the XOR of primes in X1 and
// the XOR of non primes in X2
int X1 = 1, X2 = 1;
for (int i = 0; i < n; i++) {
if (prime[arr[i]]) {
// The number is prime
X1 ^= arr[i];
}
else if (arr[i] != 1) {
// The number is non-prime
X2 ^= arr[i];
}
}
// Return the absolute difference
return abs(X1 - X2);
}
// Driver code
int main()
{
int arr[] = { 1, 3, 5, 10, 15, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Find the absolute difference
cout << calculateDifference(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Function to return
// max_element from an array
static int max_element(int[] arr)
{
int max = arr[0];
for (int ele : arr)
if (max < ele)
max = ele;
return max;
}
// Function to find the absolute difference
// between the XOR of non-primes and the
// XOR of primes in the given array
static int calculateDifference(int[] arr, int n)
{
// Find maximum value in the array
int max_val = max_element(arr);
// USE SIEVE TO FIND ALL PRIME NUMBERS
// LESS THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]".
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean[] prime = new boolean[max_val + 1];
Arrays.fill(prime, true);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * 2;
i <= max_val; i += p)
prime[i] = false;
}
}
// Store the XOR of primes in X1 and
// the XOR of non primes in X2
int x1 = 1, x2 = 1;
for (int i = 0; i < n; i++)
{
if (prime[arr[i]])
// The number is prime
x1 ^= arr[i];
else if (arr[i] != 1)
// The number is non-prime
x2 ^= arr[i];
}
// Return the absolute difference
return Math.abs(x1 - x2);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 3, 5, 10, 15, 7 };
int n = arr.length;
// Find the absolute difference
System.out.println(calculateDifference(arr, n));
}
}
// This code is contributed by
// sanjeev2552Python3
# Python3 implementation of the approach
# Function to find the absolute difference
# between the XOR of non-primes and the
# XOR of primes in the given array
def calculateDifference(arr, n):
# Find maximum value in the array
max_val = max(arr)
# USE SIEVE TO FIND ALL PRIME NUMBERS
# LESS THAN OR EQUAL TO max_val
# Create a boolean array "prime[0..n]".
# A value in prime[i] will finally be
# false if i is Not a prime, else true.
prime = [True for i in range(max_val + 1)]
# Remaining part of SIEVE
prime[0] = False
prime[1] = False
for p in range(2, max_val + 1):
if p * p > max_val + 1:
break
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(2 * p, max_val + 1, p):
prime[i] = False
# Store the XOR of primes in X1 and
# the XOR of non primes in X2
X1 = 1
X2 = 1
for i in range(n):
if (prime[arr[i]]):
# The number is prime
X1 ^= arr[i]
elif (arr[i] != 1):
# The number is non-prime
X2 ^= arr[i]
# Return the absolute difference
return abs(X1 - X2)
# Driver code
arr = [1, 3, 5, 10, 15, 7]
n = len(arr)
# Find the absolute difference
print(calculateDifference(arr, n))
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return
// max_element from an array
static int max_element(int[] arr)
{
int max = arr[0];
foreach (int ele in arr)
if (max < ele)
max = ele;
return max;
}
// Function to find the absolute difference
// between the XOR of non-primes and the
// XOR of primes in the given array
static int calculateDifference(int[] arr, int n)
{
// Find maximum value in the array
int max_val = max_element(arr);
// USE SIEVE TO FIND ALL PRIME NUMBERS
// LESS THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]".
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
bool[] prime = new bool[max_val + 1];
for(int index = 0; index < max_val + 1; index++)
prime[index] = true;
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * 2;
i <= max_val; i += p)
prime[i] = false;
}
}
// Store the XOR of primes in X1 and
// the XOR of non primes in X2
int x1 = 1, x2 = 1;
for (int i = 0; i < n; i++)
{
if (prime[arr[i]])
// The number is prime
x1 ^= arr[i];
else if (arr[i] != 1)
// The number is non-prime
x2 ^= arr[i];
}
// Return the absolute difference
return Math.Abs(x1 - x2);
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 1, 3, 5, 10, 15, 7 };
int n = arr.Length;
// Find the absolute difference
Console.WriteLine(calculateDifference(arr, n));
}
}
// This code is contributed by PrinciRaj1992输出：
4