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C Program for kth largest factor of N (TCS Codevita) | PrepInsta
kth largest factor of N Problem
Seeking for the greatest minds and talents in the country and around TCS organizes a Coding Competition each year called TCS CodeVita. Kth largest factor of N is one of the sample problem of this year TCS CodeVita season 9 competition. This problem can be solved using a counter variable and a variable holding the position for largest factor yet , here we have provided the solution for the above problem in C Language.
Problem Description
Question -: 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 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 Format: The input is a comma-separated list of positive integer pairs (N, k).
Output Format: 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.
#include <stdio.h> void main() { int number,pos_of_factor,i,c=0; scanf("%d",&number); scanf("%d",&pos_of_factor); for(i=number;i>=1;i--) { if((number%i)==0) c++; if(c==pos_of_factor) { printf("%d",i); break; } } if(c<pos_of_factor) printf("1"); }
Output 12,3 4
C++
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another way in c language
#include
int i,j,k,x,y,z,a,b,c,sum=0,flag,flag1,n,arr[10000000];
int main()
{
printf(“enter the positive numbers”);
scanf(“%d,%d”,&a,&b);
for(i=1;i<=a;i++)
{
if(a%i==0)
{
arr[j]=i;
k=j;
j++;
}
}
if(b than factors, output is %d\n”,b,1);
}
for(j=0;j<=k;j++)
{
printf(" %d\t",arr[j]);
}
return 0;
}