位数≤D的所有素数之和

给定一个整数D ，任务是找到位数小于或等于D的所有素数的总和。

例子：

Input: D = 2 
Output: 1060 
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 
47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97 are 
the prime numbers having digits less than or equal 
to 2 and the sum of these prime numbers is 1060.

Input: D = 3 
Output: 76127 
 

方法：使用埃拉托色尼筛法生成所有素数，直到最大 D 位数，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++14

// 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 d-digits
    int maxVal = pow(10, 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 = 3;
 
    cout << sumPrime(d);
 
    return 0;
}

Java

// Java implementation of the approach
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)
    {
        int i;
        // Maximum number of d-digits
        int maxVal = (int)Math.pow(10, d) - 1;
     
        // Sieve of Eratosthenes
        boolean prime[] = new boolean[maxVal + 1];
         
        for(i = 0; i < maxVal + 1; i++)
            prime[i] = true;
             
        sieve(prime, maxVal);
     
        // To store the required sum
        int sum = 0;
     
        for (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 = 3;
     
        System.out.println(sumPrime(d));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
from math import sqrt
 
# Function for Sieve of Eratosthenes
def sieve(prime, n) :
 
    prime[0] = False;
    prime[1] = False;
    for p in range(2, int(sqrt(n)) + 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 d-digits
    maxVal = (10 ** d) - 1;
 
    # Sieve of Eratosthenes
    prime = [True] * (maxVal + 1);
    sieve(prime, maxVal);
 
    # To store the required sum
    sum = 0;
 
    for i in range(2, maxVal + 1) :
 
        # If current element is prime
        if (prime[i]) :
            sum += i;
 
    return sum;
 
# Driver code
if __name__ == "__main__" :
 
    d = 3;
 
    print(sumPrime(d));
 
# This code is contributed by kanugargng

C#

// C# implementation of the above approach
using System;
     
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)
    {
        int i;
        // Maximum number of d-digits
        int maxVal = (int)Math.Pow(10, d) - 1;
     
        // Sieve of Eratosthenes
        Boolean []prime = new Boolean[maxVal + 1];
         
        for(i = 0; i < maxVal + 1; i++)
            prime[i] = true;
             
        sieve(prime, maxVal);
     
        // To store the required sum
        int sum = 0;
     
        for (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 = 3;
     
        Console.WriteLine(sumPrime(d));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
 
// Function for Sieve of Eratosthenes
function sieve(prime, n)
{
    prime[0] = false;
    prime[1] = false;
     
    for(let p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for(let i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to return the sum of
// the required prime numbers
function sumPrime(d)
{
     
    // Maximum number of d-digits
    let maxVal = Math.pow(10, d) - 1;
 
    // Sieve of Eratosthenes
    let prime = new Array(maxVal + 1);
    prime.fill(true)
    sieve(prime, maxVal);
 
    // To store the required sum
    let sum = 0;
 
    for(let i = 2; i <= maxVal; i++)
    {
         
        // If current element is prime
        if (prime[i])
        {
            sum += i;
        }
    }
    return sum;
}
 
// Driver code
let d = 3;
 
document.write(sumPrime(d));
 
// This code is contributed by _saurabh_jaiswal
 

输出：

76127