计算一个数组中的素数
给定一个由 N 个正整数组成的数组 arr[]。任务是编写一个程序来计算给定数组中质数元素的数量。 例 :
Input: arr[] = {1, 3, 4, 5, 7}
Output: 3
There are three primes, 3, 5 and 7
Input: arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: 4
天真方法:一个简单的解决方法是遍历数组并保持检查每个元素是否是素数并同时保持素数元素的计数。 高效方法:使用厄拉多塞的筛生成数组最大元素之前的所有素数，并将它们存储在散列中。现在遍历数组，使用哈希表找到那些质数元素的数量。 以下是上述方法的实施:
C++
// CPP program to find count of
// primes in given array.
#include <bits/stdc++.h>
using namespace std;
// Function to find count of prime
int primeCount(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<bool> 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;
}
}
// Find all primes in arr[]
int count = 0;
for (int i = 0; i < n; i++)
if (prime[arr[i]])
count++;
return count;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << primeCount(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
import java.util.Arrays;
import java.util.Vector;
// Java program to find count of
// primes in given array.
class GFG
{
// Function to find count of prime
static int primeCount(int arr[], int n)
{
// Find maximum value in the array
//.*max_element(arr, arr+n);
int max_val = Arrays.stream(arr).max().getAsInt();
// 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];
for (int i = 0; i < max_val + 1; i++)
{
prime[i] = 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;
}
}
}
// Find all primes in arr[]
int count = 0;
for (int i = 0; i < n; i++)
{
if (prime[arr[i]])
{
count++;
}
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
int n = arr.length;
System.out.println(primeCount(arr, n));
}
}
// This code is contributed by
// PrinciRaj1992
Python 3
# Python 3 program to find count of
# primes in given array.
from math import sqrt
# Function to find count of prime
def primeCount(arr, n):
# Find maximum value in the array
max_val = arr[0];
for i in range(len(arr)):
if(arr[i] > max_val):
max_val = arr[i]
# 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
k = int(sqrt(max_val)) + 1
for p in range(2, k, 1):
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * 2, max_val + 1, p):
prime[i] = False
# Find all primes in arr[]
count = 0
for i in range(0, n, 1):
if (prime[arr[i]]):
count += 1
return count
# Driver code
if __name__ == '__main__':
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
print(primeCount(arr, n))
# This code is contributed by
# Shashank_Sharma
C
// C# program to find count of
// primes in given array.
using System;
using System.Linq;
class GFG
{
// Function to find count of prime
static int primeCount(int []arr, int n)
{
// Find maximum value in the array
//.*max_element(arr, arr+n);
int max_val = arr.Max();
// 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];
for (int i = 0; i < max_val + 1; i++)
{
prime[i] = 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;
}
}
}
// Find all primes in arr[]
int count = 0;
for (int i = 0; i < n; i++)
{
if (prime[arr[i]])
{
count++;
}
}
return count;
}
// Driver code
public static void Main()
{
int []arr = {1, 2, 3, 4, 5, 6, 7};
int n = arr.Length;
Console.WriteLine(primeCount(arr, n));
}
}
//This code is contributed by 29AjayKumar
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find count
// of primes in given array.
// Function to find count of prime
function primeCount($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 = array_fill(0, $max_val + 1, true);
// Remaining part of SIEVE
$prime[0] = false;
$prime[1] = false;
for ($p = 2; $p * $p <= $max_val; $p++)
{
// If prime[p] is not changed,
// then it is a prime
if ($prime[$p] == true)
{
// Update all multiples of p
for ($i = $p * 2;
$i <= $max_val; $i += $p)
$prime[$i] = false;
}
}
// Find all primes in arr[]
$count = 0;
for ($i = 0; $i < $n; $i++)
if ($prime[$arr[$i]])
$count++;
return $count;
}
// Driver code
$arr = array(1, 2, 3, 4, 5, 6, 7 );
$n = sizeof($arr);
echo primeCount($arr, $n);
// This code is contributed by mits
?>
java 描述语言
<script>
// Javascript program to find count
// of primes in given array.
// Function to find count of prime
function primeCount(arr, n)
{
// Find maximum value in the array
let max_val = arr.sort((a, b) => b - a)[0];
// 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.
let prime = new Array(max_val + 1).fill(true);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (let 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 (let i = p * 2;
i <= max_val; i += p)
prime[i] = false;
}
}
// Find all primes in arr[]
let count = 0;
for (let i = 0; i < n; i++)
if (prime[arr[i]])
count++;
return count;
}
// Driver code
let arr = new Array(1, 2, 3, 4, 5, 6, 7 );
let n = arr.length;
document.write(primeCount(arr, n));
// This code is contributed by _saurabh_jaiswal
</script>
Output:
4
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