Tuesday, 7 March 2017

Geeksforgeeks Solution For " All divisors of a natural number "

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Problem :- Given a natural number n, print all distinct divisors of it including 1 and the number itself.

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Solution :-

#include <iostream>
using namespace std;
int main() 

{
int t;
cin>>t;

while(t--)
{
   int n,i,flag=0;
   cin>>n;
   for(i=1;i<=n;i++)
   {
       if(n%i==0)
       cout<<i<<" ";
   }cout<<endl;
}

return 0;
}

Output:-

Geeksforgeeks Solution For " All divisors of a natural number "

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