给定一个数组arr ，任务是找到数组中具有质数频率的元素之和。
注意： 1既不是素数也不是复合数。
例子：
Input: arr[] = {5, 4, 6, 5, 4, 6}
Output: 15
All the elements appear 2 times which is a prime
So, 5 + 4 + 6 = 15
Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}
Output: 5
Only 2 and 3 appears prime number of times i.e. 3 and 5 respectively.
So, 2 + 3 = 5
方法：
- 遍历数组并将所有元素的频率存储在地图中。
- 建立Eratosthenes筛子，该筛子将用于测试O(1)时间中数字的素数。
- 使用上一步中计算出的Sieve数组计算具有质数频率的元素之和。
下面是上述方法的实现：
C++
// C++ program to find sum of elements
// in an array having prime frequency
#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 sum of elements
// in an array having prime frequency
int sumOfElements(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]]++;
int sum = 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]) {
sum += (it->first);
}
}
return sum;
}
// Driver code
int main()
{
int arr[] = { 5, 4, 6, 5, 4, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << sumOfElements(arr, n);
return 0;
} Java
// Java program to find sum 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 sum of elements
// in an array having prime frequency
static int sumOfElements(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 sum = 0;
// Traverse the map
for (Map.Entry entry : m.entrySet())
{
int key = entry.getKey();
int value = entry.getValue();
// Count the number of elements
// having prime frequencies
if (prime[value])
{
sum += (key);
}
}
return sum;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 5, 4, 6, 5, 4, 6 };
int n = arr.length;
System.out.println(sumOfElements(arr, n));
}
}
// This code is contributed by ghanshyampandey Python3
# Python3 program to find Sum of elements
# in an array having prime frequency
import math as mt
# 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, mt.ceil(mt.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
# Function to return the Sum of elements
# in an array having prime frequency
def SumOfElements(arr, n):
prime = [True for i in range(n + 1)]
SieveOfEratosthenes(prime, n + 1)
i, j = 0, 0
# Map is used to store
# element frequencies
m = dict()
for i in range(n):
if arr[i] in m.keys():
m[arr[i]] += 1
else:
m[arr[i]] = 1
Sum = 0
# Traverse the map using iterators
for i in m:
# Count the number of elements
# having prime frequencies
if (prime[m[i]]):
Sum += (i)
return Sum
# Driver code
arr = [5, 4, 6, 5, 4, 6 ]
n = len(arr)
print(SumOfElements(arr, n))
# This code is contributed
# by Mohit kumar 29C#
// C# program to find sum 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 sum of elements
// in an array having prime frequency
static int sumOfElements(int []arr, int n)
{
bool []prime = new bool[n + 1];
for(int 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 (int i = 0 ; i < n; i++)
{
if(m.ContainsKey(arr[i]))
{
var val = m[arr[i]];
m.Remove(arr[i]);
m.Add(arr[i], val + 1);
}
else
{
m.Add(arr[i], 1);
}
}
int sum = 0;
// Traverse the map
foreach(KeyValuePair entry in m)
{
int key = entry.Key;
int value = entry.Value;
// Count the number of elements
// having prime frequencies
if (prime[value])
{
sum += (key);
}
}
return sum;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 5, 4, 6, 5, 4, 6 };
int n = arr.Length;
Console.WriteLine(sumOfElements(arr, n));
}
}
// This code is contributed by 29AjayKumar 输出：
15