Kth largest factor of N - Code Vita 2017 | round 1

 kth largest factor of N  

Problem

A positive integer d is said to be a factor of another positive integer N if when N is divided by d, the remainder obtained is zero. For example, for the number 12, there are 6 factors 1, 2, 3, 4, 6, 12. Every positive integer k has at least two factors, 1 and the number k itself.

 Given two positive integers N and k, write a program to print the kth largest factor of N.  

Input 

The input is a comma separated list of positive integer pairs (N, k)

Output 

The kth highest factor of N. If N does not have k factors, the output should be 1.

Constraints 

1<N<10000000000. 1<k<600

You can assume that N will have no prime factors which are larger than 13.

Example 1 

Input: 12,3

Output: 4

Explanation:

N is 12, k is 3. The factors of 12 are (1,2,3,4,6,12). The highest factor is 12 and the third largest factor is 4. The output must be 4

Example 2 

Input: 30,9

Output: 1

Explanation:

 N is 30, k is 9. The factors of 30 are (1,2,3,5,6,10,15,30). There are only 8 factors. As k is more than the number of factors, the output is 1.

Program

#include <stdio.h>

int main() {

int n,k,i,c=0;

scanf("%d",&n);

scanf("%d",&k);

for(i=n;i>=1;i--)

{

    if((n%i)==0)

    c++;

    if(c==k)

    {

    printf("%d",i);

    break;

    }

}

if(c!=k)

printf("1");

return 0;

}

Output

12 3

4

You can run it in an online IDE: https://ide.geeksforgeeks.org/6xxVFQA20s

Your feedback are most welcome! If you have any doubt you can contact me or leave a comment!! Cheers!!!

Related Links: The Great Chase

Consecutive Prime Sum | Code Vita 2016 | round 1

Problem Description:

Some prime numbers can be expressed as Sum of other consecutive prime numbers.
For example

5 = 2 + 3
17 = 2 + 3 + 5 + 7
41 = 2 + 3 + 5 + 7 + 11 + 13

Your 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.
Write code to find out number of prime numbers that satisfy the above mentioned property in a given range.
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.

Sample Input and Output

SNo.	Input	Output	Comment
1	20	2	(Below 20, there are 2 such numbers: 5 and 17). 5=2+3 17=2+3+5+7
2	15	1

Program:

#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;
}

Output:

20

2

You can also run it on a online IDE: https://ide.geeksforgeeks.org/XcOKzTd4Ik

If you have any doubt you can contact me or comment it below! Your comments and feedbacks are also welcomed!! Cheers!!

Related Links: Jumble with Numbers