The inner loop of this pseudocode runs only for the divisors of n for i=1 to n:   if(n%i==0):      for j=1 to n:         Do something in constant timeIf I can get a bound on the number of divisor of n(The condition in the if statement) then I can get its time complexity.Please Help ?
Tight bound on the number of divisors of n
algorithms;time complexity