Question: In the program below, assume that n > = 1 and n is a perfect cube: that is , there is z in N so

In the program below, assume that n >=1 and n is a perfect cube: that is, there is z in N so that n = z3. Find a function f (n) that counts the number of times that fnc() is called, as a function of the input n. Make sure that you are only counting calls to fnc() and not any other steps. You can use the given loop invariant without proof. You may use without proof the fact that m i=0 i3=( m i=0 i )2, for any m in Z, m >=0.1.//Pre: n as described 2. a03. k14.//LoopInv: k ==(a+1)*(a+1)*(a+1)5. while(k <= n): 6. j07. while(j < k): 8. fnc()9. jj+110. aa+111. kk+(3*a*a)+(3*a)+13

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Programming Questions!