检查数字是否可以写为’k’质数的总和

给定两个数字N和K。我们需要确定“ N”是否可以写为“ K”素数的总和。
给定N <= 10 ^ 9
例子 ：

Input  : N = 10 K = 2
Output : Yes 
        10 can be written as 5 + 5

Input  : N = 2 K = 2
Output : No

这个想法是使用戈德巴赫猜想的，该猜想说每个偶数整数(大于2)都可以表示为两个素数之和。
如果N = 2K且K = 1 ：答案是肯定的，如果N是质数
如果N> = 2K且K = 2 ：如果N是偶数，则答案为是(戈德巴赫猜想)，如果N为奇数，则当N-2不是素数时答案为否，如果N-2是素数，则答案为Yes。质数。这是因为我们知道奇数+奇数=偶数和偶数+奇数=奇数。因此，当N为奇数且K = 2时，一个数必须为2，因为它是唯一的偶数质数，因此现在的答案取决于N-2是否为奇数。
如果N> = 2K并且K> = 3：答案将始终是。当N是偶数N – 2 *(K-2)也是偶数时，N – 2 *(K – 2)可以写成两个质数之和(哥德巴赫猜想)，p和q可以写成2， 2…..K – 2次，p，q。当N为奇数时，N – 3 -2 *(K – 3)为偶数，因此可以将其写为两个质数p的和，q和N可以写为2、2…..K-3倍，3， ，

C++

// C++ implementation to check if N can be
// written as sum of k primes
#include
using namespace std;
 
// Checking if a number is prime or not
bool isprime(int x)
{
   
    // check for numbers from 2 to sqrt(x)
    // if it is divisible return false
    for (int i = 2; i * i <= x; i++)
        if (x % i == 0)
            return false;
    return true;
}
 
// Returns true if N can be written as sum
// of K primes
bool isSumOfKprimes(int N, int K)
{
    // N < 2K directly return false
    if (N < 2*K)
        return false;
 
    // If K = 1 return value depends on primality of N
    if (K == 1)
        return isprime(N);
 
    if (K == 2)
    {
        // if N is even directly return true;
        if (N % 2 == 0)
            return true;
 
        // If N is odd, then one prime must
        // be 2. All other primes are odd
        // and cannot have a pair sum as even.
        return isprime(N - 2);
    }
 
    // If K >= 3 return true;
    return true;
}
 
// Driver function
int main()
{
    int n = 10, k = 2;
    if (isSumOfKprimes (n, k))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java implementation to check if N can be
// written as sum of k primes
public class Prime
{
    // Checking if a number is prime or not
    static boolean isprime(int x)
    {
        // check for numbers from 2 to sqrt(x)
        // if it is divisible return false
        for (int i=2; i*i<=x; i++)
            if (x%i == 0)
             
                return false;
        return true;
    }
     
    // Returns true if N can be written as sum
    // of K primes
    static boolean isSumOfKprimes(int N, int K)
    {
        // N < 2K directly return false
        if (N < 2*K)
            return false;
         
        // If K = 1 return value depends on primality of N
        if (K == 1)
            return isprime(N);
             
        if (K == 2)
        {
            // if N is even directly return true;
            if (N%2 == 0)
                return true;
                 
            // If N is odd, then one prime must
            // be 2. All other primes are odd
            // and cannot have a pair sum as even.
            return isprime(N - 2);
        }
         
        // If K >= 3 return true;
        return true;
    }
     
    public static void main (String[] args)
    {
        int n = 10, k = 2;
        if (isSumOfKprimes (n, k))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}
// Contributed by Saket Kumar

Python3

# Python implementation to check
# if N can be written as sum of
# k primes
 
# Checking if a number is prime
# or not
def isprime(x):
     
    # check for numbers from 2
    # to sqrt(x) if it is divisible
    # return false
    i = 2
    while(i * i <= x):
        if (x % i == 0):
            return 0
    return 1
 
# Returns true if N can be written
# as sum of K primes
def isSumOfKprimes(N, K):
     
    # N < 2K directly return false
    if (N < 2 * K):
        return 0
 
    # If K = 1 return value depends
    # on primality of N
    if (K == 1):
        return isprime(N)
 
    if (K == 2):
         
        # if N is even directly
        # return true;
        if (N % 2 == 0):
            return 1
 
        # If N is odd, then one
        # prime must be 2. All
        # other primes are odd
        # and cannot have a pair
        # sum as even.
        return isprime(N - 2);
     
 
    # If K >= 3 return true;
    return 1
 
# Driver function
n = 10
k = 2
if (isSumOfKprimes (n, k)):
    print ("Yes")
else:
    print ("No")
 
# This code is Contributed by Sam007.

C#

// C# implementation to check if N can be
// written as sum of k primes
using System;
         
class GFG {
     
    // Checking if a number is prime or not
    static bool isprime(int x)
    {
        // check for numbers from 2 to sqrt(x)
        // if it is divisible return false
        for (int i = 2; i * i <= x; i++)
            if (x % i == 0)
             
                return false;
        return true;
    }
     
    // Returns true if N can be written as sum
    // of K primes
    static bool isSumOfKprimes(int N, int K)
    {
        // N < 2K directly return false
        if (N < 2 * K)
            return false;
         
        // If K = 1 return value depends on primality of N
        if (K == 1)
            return isprime(N);
             
        if (K == 2)
        {
            // if N is even directly return true;
            if (N % 2 == 0)
                return true;
                 
            // If N is odd, then one prime must
            // be 2. All other primes are odd
            // and cannot have a pair sum as even.
            return isprime(N - 2);
        }
         
        // If K >= 3 return true;
        return true;
    }
     
    // Driver function
    public static void Main ()
    {
        int n = 10, k = 2;
        if (isSumOfKprimes (n, k))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
 
// This code is contributed by Sam007

PHP

= 3 return true;
    return true;
}
 
// Driver Code
$n = 10; $k = 2;
if (isSumOfKprimes ($n, $k))
    echo "Yes";
else
    echo"No" ;
 
// This code is contributed by vt
?>

Javascript

// javascript implementation to check if N can be
// written as sum of k primes
 
    // Checking if a number is prime or not
    function isprime(x)
    {
     
        // check for numbers from 2 to sqrt(x)
        // if it is divisible return false
        for (i = 2; i * i <= x; i++)
            if (x % i == 0)
 
                return false;
        return true;
    }
 
    // Returns true if N can be written as sum
    // of K primes
    function isSumOfKprimes(N, K)
    {
     
        // N < 2K directly return false
        if (N < 2 * K)
            return false;
 
        // If K = 1 return value depends on primality of N
        if (K == 1)
            return isprime(N);
 
        if (K == 2)
        {
         
            // if N is even directly return true;
            if (N % 2 == 0)
                return true;
 
            // If N is odd, then one prime must
            // be 2. All other primes are odd
            // and cannot have a pair sum as even.
            return isprime(N - 2);
        }
 
        // If K >= 3 return true;
        return true;
    }
 
    // Driver code
        var n = 10, k = 2;
        if (isSumOfKprimes(n, k))
            document.write("Yes");
        else
            document.write("No");
 
// This code is contributed by gauravrajput1

输出 ：

Yes