设置位最大位置≤D的所有素数之和

给定一个整数D ，任务是找到所有素数的总和，这些素数的最大置位比特位置(距离右侧最远的置位比特)小于或等于D。
注意：二进制中的2为10，最大设置位位置为2。二进制中的7为111，最大设置位位置为3。

例子：

Input: D = 3
Output: 17
2, 3, 5 and 7 are the only primes
which satisfy the given condition.

Input: D = 8
Output: 6081

方法：满足给定条件的最大数量为2 ^D – 1 。因此，使用直到2 ^D – 1的Eratosthenes筛网生成所有素数，然后找到同一范围内所有素数的总和。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function for Sieve of Eratosthenes
void sieve(bool prime[], int n)
{
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= n; p++) {
        if (prime[p] == true) {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return the sum of
// the required prime numbers
int sumPrime(int d)
{
  
    // Maximum number of the required range
    int maxVal = pow(2, d) - 1;
  
    // Sieve of Eratosthenes
    bool prime[maxVal + 1];
    memset(prime, true, sizeof(prime));
    sieve(prime, maxVal);
  
    // To store the required sum
    int sum = 0;
  
    for (int i = 2; i <= maxVal; i++) {
  
        // If current element is prime
        if (prime[i]) {
            sum += i;
        }
    }
  
    return sum;
}
  
// Driver code
int main()
{
    int d = 8;
  
    cout << sumPrime(d);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// Function for Sieve of Eratosthenes
static void sieve(boolean prime[], int n)
{
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p;
                     i <= n; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return the sum of
// the required prime numbers
static int sumPrime(int d)
{
  
    // Maximum number of the required range
    int maxVal = (int) (Math.pow(2, d) - 1);
  
    // Sieve of Eratosthenes
    boolean []prime = new boolean[maxVal + 1];
    Arrays.fill(prime, true);
    sieve(prime, maxVal);
  
    // To store the required sum
    int sum = 0;
  
    for (int i = 2; i <= maxVal; i++)
    {
  
        // If current element is prime
        if (prime[i]) 
        {
            sum += i;
        }
    }
    return sum;
}
  
// Driver code
public static void main(String[] args) 
{
    int d = 8;
  
    System.out.println(sumPrime(d));
}
}
  
// This code is contributed by PrinciRaj1992

Python 3

# Python 3 implementation of the approach
from math import sqrt, pow
  
# Function for Sieve of Eratosthenes
def sieve(prime, n):
    prime[0] = False
    prime[1] = False
    for p in range(2, int(sqrt(n)) + 1, 1):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
  
# Function to return the sum of
# the required prime numbers
def sumPrime(d):
      
    # Maximum number of the required range
    maxVal = int(pow(2, d)) - 1;
  
    # Sieve of Eratosthenes
    prime = [True for i in range(maxVal + 1)]
      
    sieve(prime, maxVal)
  
    # To store the required sum
    sum = 0
  
    for i in range(2, maxVal + 1, 1):
          
        # If current element is prime
        if (prime[i]):
            sum += i
  
    return sum
  
# Driver code
if __name__ == '__main__':
    d = 8
  
    print(sumPrime(d))
  
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the approach
using System;
using System.Linq;
  
class GFG 
{
  
// Function for Sieve of Eratosthenes
static void sieve(Boolean []prime, int n)
{
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p;
                    i <= n; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return the sum of
// the required prime numbers
static int sumPrime(int d)
{
  
    // Maximum number of the required range
    int maxVal = (int) (Math.Pow(2, d) - 1);
  
    // Sieve of Eratosthenes
    Boolean []prime = new Boolean[maxVal + 1];
      
    for (int i = 0; i <= maxVal; i++)
        prime.SetValue(true,i);
    sieve(prime, maxVal);
  
    // To store the required sum
    int sum = 0;
  
    for (int i = 2; i <= maxVal; i++)
    {
  
        // If current element is prime
        if (prime[i]) 
        {
            sum += i;
        }
    }
    return sum;
}
  
// Driver code
public static void Main(String[] args) 
{
    int d = 8;
  
    Console.WriteLine(sumPrime(d));
}
}
  
// This code is contributed by 29AjayKumar

输出：

6081