最多 N 的有趣素数

给定一个数字N ，任务是找到小于等于N的有趣素数的数量。
一个有趣的素数是任何可以写成a ² + b ^{4 的}素数，其中a和b是正整数。例如，最小的有趣素数是2 = 1 ² + 1 ⁴ 。
例子：

Input: N = 10 
Output: 2 
2 = 1² + 1⁴ 
5 = 2² + 1⁴ 
Both are interesting primes less than equal to 10
Input: N = 1000 
Output: 28 
 

天真的方法：

1. 遍历从1到N 的所有数字。
2. 对于每个数字，检查它是否是质数。
3. 如果它是素数，则检查它是否可以表示为a ² + b ⁴ ：
   • 迭代b 的所有可能值，从 1 到 N ^1/4 。
   • 对于b 的每个值，检查N – b ⁴是否是完全平方数（即它可以是²或不是）。

下面是上述方法的实现：

C++

// C++ program to find the number
// of interesting primes up to N
 
#include 
using namespace std;
 
// Function to check if a number
// is prime or not
bool isPrime(int n)
{
 
    int flag = 1;
 
    // If n is divisible by any
    // number between 2 and sqrt(n),
    // it is not prime
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            flag = 0;
            break;
        }
    }
 
    return (flag == 1 ? true : false);
}
 
// Function to check if a number
// is perfet square or not
bool isPerfectSquare(int x)
{
    // Find floating point value of
    // square root of x.
    long double sr = sqrt(x);
 
    // If square root is an integer
    return ((sr - floor(sr)) == 0);
}
 
// Function to find the number of interesting
// primes less than equal to N.
int countInterestingPrimes(int n)
{
 
    int answer = 0;
    for (int i = 2; i <= n; i++) {
 
        // Check whether the number
        // is prime or not
        if (isPrime(i)) {
 
            // Iterate for values of b
            for (int j = 1;
                 j * j * j * j <= i;
                 j++) {
 
                // Check condition for a
                if (
                    isPerfectSquare(
                        i - j * j * j * j)) {
                    answer++;
                    break;
                }
            }
        }
    }
 
    // Return the required answer
    return answer;
}
 
// Driver code
int main()
{
    int N = 10;
 
    cout << countInterestingPrimes(N);
 
    return 0;
}

Java

// Java program to find the number
// of interesting primes up to N
class GFG{
  
// Function to check if a number
// is prime or not
static boolean isPrime(int n)
{
  
    int flag = 1;
  
    // If n is divisible by any
    // number between 2 and Math.sqrt(n),
    // it is not prime
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            flag = 0;
            break;
        }
    }
  
    return (flag == 1 ? true : false);
}
  
// Function to check if a number
// is perfet square or not
static boolean isPerfectSquare(int x)
{
    // Find floating point value of
    // square root of x.
    double sr = Math.sqrt(x);
  
    // If square root is an integer
    return ((sr - Math.floor(sr)) == 0);
}
  
// Function to find the number of interesting
// primes less than equal to N.
static int countInterestingPrimes(int n)
{
  
    int answer = 0;
    for (int i = 2; i <= n; i++) {
  
        // Check whether the number
        // is prime or not
        if (isPrime(i)) {
  
            // Iterate for values of b
            for (int j = 1;
                 j * j * j * j <= i;
                 j++) {
  
                // Check condition for a
                if (
                    isPerfectSquare(
                        i - j * j * j * j)) {
                    answer++;
                    break;
                }
            }
        }
    }
  
    // Return the required answer
    return answer;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 10;
  
    System.out.print(countInterestingPrimes(N));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program to find the number
# of interesting primes up to N
import math
 
# Function to check if a number
# is prime or not
def isPrime(n):
 
    flag = 1
 
    # If n is divisible by any
    # number between 2 and sqrt(n),
    # it is not prime
    i = 2
    while(i * i <= n):
        if (n % i == 0):
            flag = 0
            break
        i += 1
     
    return (True if flag == 1 else False)
 
# Function to check if a number
# is perfet square or not
def isPerfectSquare(x):
 
    # Find floating povalue of
    # square root of x.
    sr = math.sqrt(x)
 
    # If square root is an integer
    return ((sr - math.floor(sr)) == 0)
 
# Function to find the number of interesting
# primes less than equal to N.
def countInterestingPrimes(n):
 
    answer = 0
    for i in range(2, n):
 
        # Check whether the number
        # is prime or not
        if (isPrime(i)):
 
            # Iterate for values of b
            j = 1
            while(j * j * j * j <= i):
 
                # Check condition for a
                if (isPerfectSquare(i - j * j *
                                        j * j)):
                    answer += 1
                    break
                j += 1
 
    # Return the required answer
    return answer
 
# Driver code
if __name__=='__main__':
 
    N = 10
 
    print(countInterestingPrimes(N))
 
# This code is contributed by AbhiThakur

C#

// C# program to find the number
// of interesting primes up to N
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to check if a number
// is prime or not
static bool isPrime(int n)
{
   
    int flag = 1;
   
    // If n is divisible by any
    // number between 2 and Math.Sqrt(n),
    // it is not prime
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            flag = 0;
            break;
        }
    }
   
    return (flag == 1 ? true : false);
}
   
// Function to check if a number
// is perfet square or not
static bool isPerfectSquare(int x)
{
    // Find floating point value of
    // square root of x.
    double sr = Math.Sqrt(x);
   
    // If square root is an integer
    return ((sr - Math.Floor(sr)) == 0);
}
   
// Function to find the number of interesting
// primes less than equal to N.
static int countInterestingPrimes(int n)
{
   
    int answer = 0;
    for (int i = 2; i <= n; i++) {
   
        // Check whether the number
        // is prime or not
        if (isPrime(i)) {
   
            // Iterate for values of b
            for (int j = 1;
                 j * j * j * j <= i;
                 j++) {
   
                // Check condition for a
                if (
                    isPerfectSquare(
                        i - j * j * j * j)) {
                    answer++;
                    break;
                }
            }
        }
    }
   
    // Return the required answer
    return answer;
}
   
// Driver code
public static void Main(String[] args)
{
    int N = 10;
   
    Console.Write(countInterestingPrimes(N));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// Java  script program to find the number
// of interesting primes up to N
 
// Function to check if a number
// is prime or not
function isPrime( n)
{
 
    let flag = 1;
 
    // If n is divisible by any
    // number between 2 and Math.sqrt(n),
    // it is not prime
    for (let i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            flag = 0;
            break;
        }
    }
 
    return (flag == 1 ? true : false);
}
 
// Function to check if a number
// is perfet square or not
function isPerfectSquare( x)
{
    // Find floating point value of
    // square root of x.
    let sr = Math.sqrt(x);
 
    // If square root is an integer
    return ((sr - Math.floor(sr)) == 0);
}
 
// Function to find the number of interesting
// primes less than equal to N.
function countInterestingPrimes( n)
{
 
    let answer = 0;
    for (let i = 2; i <= n; i++) {
 
        // Check whether the number
        // is prime or not
        if (isPrime(i)) {
 
            // Iterate for values of b
            for (let j = 1;
                j * j * j * j <= i;
                j++) {
 
                // Check condition for a
                if (
                    isPerfectSquare(
                        i - j * j * j * j)) {
                    answer++;
                    break;
                }
            }
        }
    }
 
    // Return the required answer
    return answer;
}
 
// Driver code
 
    let N = 10;
 
    document.write(countInterestingPrimes(N));
 
 
 
// This code is contributed by Bobby

输出：

2

时间复杂度：O(N)
有效的方法：

1. 如果我们将所有完美平方和完美四元组存储到N ，那么我们可以遍历所有对并检查结果是否为质数。
2. 为了进一步优化，我们可以使用埃拉托色尼筛将所有素数存储到N并在O(1) 中进行素数检查。

下面是上述方法的实现：

C++

// C++ program to find the number
// of interesting primes up to N.
 
#include 
using namespace std;
 
// Function to find all prime numbers
void SieveOfEratosthenes(
    int n,
    unordered_set& allPrimes)
{
    // Create a boolean array "prime[0..n]"
    // and initialize all entries as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime.
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
 
    for (int p = 2; p * p <= n; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
 
    // Store all prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            allPrimes.insert(p);
}
 
// Function to check if a number
// is perfet square or not
int countInterestingPrimes(int n)
{
    // To store all primes
    unordered_set allPrimes;
 
    SieveOfEratosthenes(n, allPrimes);
 
    // To store all interseting primes
    unordered_set intersetingPrimes;
 
    vector squares, quadruples;
 
    // Store all perfect squares
    for (int i = 1; i * i <= n; i++) {
        squares.push_back(i * i);
    }
 
    // Store all perfect quadruples
    for (int i = 1; i * i * i * i <= n; i++) {
        quadruples.push_back(i * i * i * i);
    }
 
    // Store all interseting primes
    for (auto a : squares) {
        for (auto b : quadruples) {
            if (allPrimes.count(a + b))
                intersetingPrimes.insert(a + b);
        }
    }
 
    // Return count of interseting primes
    return intersetingPrimes.size();
}
 
// Driver code
int main()
{
    int N = 10;
 
    cout << countInterestingPrimes(N);
 
    return 0;
}

Java

// Java program to find the number
// of interesting primes up to N.
import java.util.*;
 
class GFG{
  
// Function to find all prime numbers
static void SieveOfEratosthenes(
    int n, HashSet allPrimes)
{
    // Create a boolean array "prime[0..n]"
    // and initialize all entries as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime.
    boolean []prime = new boolean[n + 1];
    Arrays.fill(prime, true);
  
    for (int p = 2; p * p <= n; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
  
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    // Store all prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            allPrimes.add(p);
}
  
// Function to check if a number
// is perfet square or not
static int countInterestingPrimes(int n)
{
    // To store all primes
    HashSet allPrimes = new HashSet();
  
    SieveOfEratosthenes(n, allPrimes);
  
    // To store all interseting primes
    HashSet intersetingPrimes = new HashSet();
  
    Vector squares = new Vector()
            , quadruples = new Vector();
  
    // Store all perfect squares
    for (int i = 1; i * i <= n; i++) {
        squares.add(i * i);
    }
  
    // Store all perfect quadruples
    for (int i = 1; i * i * i * i <= n; i++) {
        quadruples.add(i * i * i * i);
    }
  
    // Store all interseting primes
    for (int a : squares) {
        for (int b : quadruples) {
            if (allPrimes.contains(a + b))
                intersetingPrimes.add(a + b);
        }
    }
  
    // Return count of interseting primes
    return intersetingPrimes.size();
}
  
// Driver code
public static void main(String[] args)
{
    int N = 10;
  
    System.out.print(countInterestingPrimes(N));
}
}
 
// This code is contributed by 29AjayKumar

蟒蛇3

# Python3 program to find the number
# of interesting primes up to N.
 
# Function to find all prime numbers
def SieveOfEratosthenes(n, allPrimes):
     
    # Create a boolean array "prime[0..n]"
    # and initialize all entries as true.
    # A value in prime[i] will finally
    # be false if i is Not a prime.
    prime = [True] * (n + 1)
     
    p = 2
    while p * p <= n:
         
        # If prime[p] is not changed,
        # then it is a prime
        if prime[p] == True:
             
            # Update all multiples of p
            # greater than or equal to
            # the square of it
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
     
    # Store all prime numbers
    for p in range(2, n + 1):
        if prime[p]:
            allPrimes.add(p)
 
# Function to check if a number
# is perfet square or not
def countInterestingPrimes(n):
     
    # To store all primes
    allPrimes = set()
     
    # To store all interseting primes
    SieveOfEratosthenes(n, allPrimes)
     
    # To store all interseting primes
    interestingPrimes = set()
     
    squares, quadruples = [], []
     
    # Store all perfect squares
    i = 1
    while i * i <= n:
        squares.append(i * i)
        i += 1
     
    # Store all perfect quadruples
    i = 1
    while i * i * i * i <= n:
        quadruples.append(i * i * i * i)
        i += 1
     
    # Store all interseting primes
    for a in squares:
        for b in quadruples:
            if a + b in allPrimes:
                interestingPrimes.add(a + b)
                 
    # Return count of interseting primes
    return len(interestingPrimes)
 
# Driver code
N = 10
print(countInterestingPrimes(N))
 
# This code is contributed by Shivam Singh

C＃

// C# program to find the number
// of interesting primes up to N.
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to find all prime numbers
static void SieveOfEratosthenes(
    int n, HashSet allPrimes)
{
    // Create a bool array "prime[0..n]"
    // and initialize all entries as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime.
    bool []prime = new bool[n + 1];
    for(int i = 0; i < n + 1; i++)
        prime[i] =  true;
   
    for (int p = 2; p * p <= n; p++) {
   
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
   
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
   
    // Store all prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            allPrimes.Add(p);
}
   
// Function to check if a number
// is perfet square or not
static int countInterestingPrimes(int n)
{
    // To store all primes
    HashSet allPrimes = new HashSet();
   
    SieveOfEratosthenes(n, allPrimes);
   
    // To store all interseting primes
    HashSet intersetingPrimes = new HashSet();
   
    List squares = new List()
            , quadruples = new List();
   
    // Store all perfect squares
    for (int i = 1; i * i <= n; i++) {
        squares.Add(i * i);
    }
   
    // Store all perfect quadruples
    for (int i = 1; i * i * i * i <= n; i++) {
        quadruples.Add(i * i * i * i);
    }
   
    // Store all interseting primes
    foreach (int a in squares) {
        foreach (int b in quadruples) {
            if (allPrimes.Contains(a + b))
                intersetingPrimes.Add(a + b);
        }
    }
   
    // Return count of interseting primes
    return intersetingPrimes.Count;
}
   
// Driver code
public static void Main(String[] args)
{
    int N = 10;
   
    Console.Write(countInterestingPrimes(N));
}
}
  
// This code is contributed by Rajput-Ji

输出：

2