Coding Sample Questions | Part 1 | Codewindow.in

Problem Statement :

Some prime numbers can be expressed as a sum of other
consecutive prime numbers. For example 5 = 2 + 3, 17 = 2 + 3 + 5 + 7, 41 = 2 + 3 + 5 + 7 + 11 + 13. Now the task is to find
out how many prime numbers which satisfy this property are
present in the range 3 to N subject to a constraint that
summation should always start with number 2.

Input Format:
First line contains a number N
Output Format:
Print the total number of all such prime numbers which are
less than or equal to N.
Constraints:
2<N<=12,000,000,000

Sample Input 1 :
20
Sample Output 1 :
2
Explanation :
5 = 2+3
17 = 2+3+5+7
Sample Input 2 :
15
Sample Output 2 :
1
Explanation :
5 = 2+3

Solution:

C

//www.codewindow.in
//start
#include <stdio.h>
int prime(int b)
{
int j,cnt;
cnt=1;
    for(j=2;j<=b/2;j++)
    {
    if(b%j==0)
    cnt=0;
    }
if(cnt==0)
    return 1;
else
    return 0;
}
int main() {
    int i,j,n,cnt,a[25],c,sum=0,count=0,k=0;
    scanf("%d",&n);
    for(i=2;i<=n;i++)
    {
    cnt=1;
    for(j=2;j<=n/2;j++)
    {
        if(i%j==0)
        cnt=0;
    }
        if(cnt==1)
        {
        a[k]=i;
        k++;
        }
    }
    for(i=0;i<k;i++)
    {
        sum=sum+a[i];
        c= prime(sum);
    if(c==1)
        count++;
    }
    printf("%d",count);
    
    return 0;
    }
//end

JAVA

//www.codewindow.in
//start
import java.util.Scanner;
class Main {
static int prime(int b) {
    int j,cnt;
    cnt=1;
    
    for (j = 2; j <= b/2; j++) 
    {
    if(b%j==0)
    cnt=0;
    }
    
    if(cnt==0)
        return 1;
    else
        return 0;
    }
    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    
    int i,j,n=0,cnt,c=0,sum=0,count=0,k=0;
    
    Main t = new Main();
    
    int[] a = new int[25];
    System.out.println("Enter no");
    n = sc.nextInt();
    
    for (i = 2; i <=n ; i++) 
    {
    cnt=1;
    for (j = 2; j <= n/2; j++) 
        {
        if(i%j==0)
        cnt=0;
        }
        
        if(cnt==1) 
        {
        a[k]=i;
        k++;
        }
    
        
    }
    
    for (i = 0; i < k; i++) 
    {
        sum=sum+a[i];
        c=t.prime(sum);
        if(c==1)
            count++;
    }
    System.out.println(count);
    }
}

//end

Python

#www.codewindow.in
#please follow the indentation as its a must in python programing

num = int(input())
arr = []
sum = 0
count = 0
if num > 1:
    for i in range(2, num + 2):
        for j in range(2, i):
            if i % j == 0:
                break
        else:
            arr.append(i)
def is_prime(sum):
    for i in range(2, (sum // 2) +2):
        if sum % i == 0:
            return False
        else:
            return True
for i in range(0, len(arr)):
    sum = sum + arr[i]
    if sum <= num:
        if is_prime(sum):
            count = count + 1
print(count)

#end

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