找到第三个数字，使所有三个数字的和成为质数

给定两个数字A和B。任务是找到大于或等于1的最小正整数，以使所有三个数字的和成为质数。
例子：

Input: A = 2, B = 3 
Output: 2 
Explaination: 
The third number is 2 because if we add all three number the 
sum obtained is 7 which is a prime number.
Input: A = 5, B = 3 
Output: 3 
 

方法：

1. 首先，将给定2个数字的和存储在sum变量中。
2. 现在，检查sum和1的按位与(＆)是否等于1(检查数字是否为奇数)。
3. 如果等于1，则将2分配给变量temp，然后转到步骤4。
4. 否则检查sum和temp变量的和值是否为质数。如果是质数，则打印temp变量的值，否则将2添加到temp变量，直到它小于质数。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function that will check
// whether number is prime or not
bool prime(int n)
{
    for (int i = 2; i * i <= n; i++) {
 
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to print the 3rd number
void thirdNumber(int a, int b)
{
    int sum = 0, temp = 0;
 
    sum = a + b;
    temp = 1;
 
    // If the sum is odd
    if (sum & 1) {
        temp = 2;
    }
 
    // If sum is not prime
    while (!prime(sum + temp)) {
        temp += 2;
    }
 
    cout << temp;
}
 
// Driver code
int main()
{
    int a = 3, b = 5;
 
    thirdNumber(a, b);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
     
class GFG
{
     
// Function that will check
// whether number is prime or not
static boolean prime(int n)
{
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to print the 3rd number
static void thirdNumber(int a, int b)
{
    int sum = 0, temp = 0;
 
    sum = a + b;
    temp = 1;
 
    // If the sum is odd
    if (sum == 0)
    {
        temp = 2;
    }
 
    // If sum is not prime
    while (!prime(sum + temp))
    {
        temp += 2;
    }
    System.out.print(temp);
}
 
// Driver code
static public void main (String []arr)
{
    int a = 3, b = 5;
     
    thirdNumber(a, b);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the above approach
 
# Function that will check
# whether number is prime or not
def prime(n):
    for i in range(2, n + 1):
        if i * i > n + 1:
            break
 
        if (n % i == 0):
            return False
 
    return True
 
# Function to print the 3rd number
def thirdNumber(a, b):
    summ = 0
    temp = 0
 
    summ = a + b
    temp = 1
 
    #If the summ is odd
    if (summ & 1):
        temp = 2
 
    #If summ is not prime
    while (prime(summ + temp) == False):
        temp += 2
 
    print(temp)
 
# Driver code
a = 3
b = 5
 
thirdNumber(a, b)
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the above approach
using System;
 
class GFG
{
     
// Function that will check
// whether number is prime or not
static bool prime(int n)
{
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to print the 3rd number
static void thirdNumber(int a, int b)
{
    int sum = 0, temp = 0;
 
    sum = a + b;
    temp = 1;
 
    // If the sum is odd
    if (sum == 0)
    {
        temp = 2;
    }
 
    // If sum is not prime
    while (!prime(sum + temp))
    {
        temp += 2;
    }
    Console.Write (temp);
}
 
// Driver code
static public void Main ()
{
    int a = 3, b = 5;
     
    thirdNumber(a, b);
}
}
 
// This code is contributed by Sachin.

Javascript

    // Javascript implementation of the above approach
     
    // Function that will check
    // whether number is prime or not
    function prime(n)
    {
        for (let i = 2; i * i <= n; i++) {
 
            if (n % i == 0)
                return false;
        }
        return true;
    }
 
    // Function to print the 3rd number
    function thirdNumber(a, b)
    {
        let sum = 0, temp = 0;
 
        sum = a + b;
        temp = 1;
 
        // If the sum is odd
        if ((sum & 1) != 0) {
            temp = 2;
        }
 
        // If sum is not prime
        while (!prime(sum + temp)) {
            temp += 2;
        }
        document.write(temp);
    }
     
    let a = 3, b = 5;
    thirdNumber(a, b);
 
// This code is contributed by divyeshrabadiya07.

输出：

3