给定一个包含N 个元素的数组arr[] ,任务是找到数组中具有素数频率的元素的异或。请注意, 1既不是质数也不是合数。
例子:
Input: arr[] = {5, 4, 6, 5, 4, 6}
Output: 7
All the elements appear 2 times which is a prime
So, 5 ^ 4 ^ 6 = 7
Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}
Output: 1
Only 2 and 3 appears prime number of times i.e. 3 and 5 respectively.
So, 2 ^ 3 = 1
方法:
- 遍历数组并将所有元素的频率存储在地图中。
- 构建 Eratosthenes 筛,用于在 O(1) 时间内测试数字的素性。
- 使用上一步计算的 Sieve 数组计算具有素数频率的元素的异或。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to create Sieve to check primes
void SieveOfEratosthenes(bool prime[], int p_size)
{
// False here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function to return the xor of elements
// in an array having prime frequency
int xorPrimeFreq(int arr[], int n)
{
bool prime[n + 1];
memset(prime, true, sizeof(prime));
SieveOfEratosthenes(prime, n + 1);
int i, j;
// Map is used to store
// element frequencies
unordered_map m;
for (i = 0; i < n; i++)
m[arr[i]]++;
long xorVal = 0;
// Traverse the map using iterators
for (auto it = m.begin(); it != m.end(); it++) {
// Count the number of elements
// having prime frequencies
if (prime[it->second]) {
xorVal ^= it->first;
}
}
return xorVal;
}
// Driver code
int main()
{
int arr[] = { 5, 4, 6, 5, 4, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << xorPrimeFreq(arr, n);
return 0;
}
Java
// Java program to find xor of elements
// in an array having prime frequency
import java.util.*;
class GFG
{
// Function to create Sieve to check primes
static void SieveOfEratosthenes(boolean prime[],
int p_size)
{
// False here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2;
i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function to return the xor of elements
// in an array having prime frequency
static int xorOfElements(int arr[], int n)
{
boolean prime[] = new boolean[n + 1];
Arrays.fill(prime, true);
SieveOfEratosthenes(prime, n + 1);
int i, j;
// Map is used to store
// element frequencies
HashMap m = new HashMap<>();
for (i = 0; i < n; i++)
{
if(m.containsKey(arr[i]))
m.put(arr[i], m.get(arr[i]) + 1);
else
m.put(arr[i], 1);
}
int xor = 0;
// Traverse the map
for (Map.Entry entry : m.entrySet())
{
int key = entry.getKey();
int value = entry.getValue();
// xor the elements
// having prime frequencies
if (prime[value])
{
xor ^= (key);
}
}
return xor;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 5, 4, 6, 5, 4, 6 };
int n = arr.length;
System.out.println(xorOfElements(arr, n));
}
}
// This code is contributed by NikhilRathor
Python3
# Python3 implementation of the approach
from math import sqrt
# Function to create Sieve to check primes
def SieveOfEratosthenes(prime, p_size) :
# False here indicates
# that it is not prime
prime[0] = False;
prime[1] = False;
for p in range(2, int(sqrt(p_size)) + 1) :
# If prime[p] is not changed,
# then it is a prime
if (prime[p]) :
# Update all multiples of p,
# set them to non-prime
for i in range(p * 2, p_size + 1, p) :
prime[i] = False;
return prime
# Function to return the xor of elements
# in an array having prime frequency
def xorPrimeFreq( arr, n) :
prime = [True] * (n + 1);
prime = SieveOfEratosthenes(prime, n + 1);
# Map is used to store
# element frequencies
m = dict.fromkeys(arr, 0);
for i in range(n) :
m[arr[i]] += 1;
xorVal = 0;
# Traverse the map using iterators
for key,value in m.items() :
# Count the number of elements
# having prime frequencies
if (prime[value]) :
xorVal ^= key;
return xorVal;
# Driver code
if __name__ == "__main__" :
arr = [ 5, 4, 6, 5, 4, 6 ];
n = len(arr);
print(xorPrimeFreq(arr, n));
# This code is contributed by AnkitRai01
C#
// C# program to find xor of elements
// in an array having prime frequency
using System;
using System.Collections.Generic;
class GFG
{
// Function to create Sieve to check primes
static void SieveOfEratosthenes(bool []prime,
int p_size)
{
// False here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2;
i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function to return the xor of elements
// in an array having prime frequency
static int xorOfElements(int []arr, int n)
{
int i, j;
bool []prime = new bool[n + 1];
for(i = 0; i< n + 1; i++)
prime[i] = true;
SieveOfEratosthenes(prime, n + 1);
// Map is used to store
// element frequencies
Dictionary m = new Dictionary();
for (i = 0; i < n; i++)
{
if(m.ContainsKey(arr[i]))
m[arr[i]] = m[arr[i]] + 1;
else
m.Add(arr[i], 1);
}
int xor = 0;
// Traverse the map
foreach(KeyValuePair entry in m)
{
int key = entry.Key;
int value = entry.Value;
// xor the elements
// having prime frequencies
if (prime[value])
{
xor ^= (key);
}
}
return xor;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 5, 4, 6, 5, 4, 6 };
int n = arr.Length;
Console.WriteLine(xorOfElements(arr, n));
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
7
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