Vyasa has to complete a programming assignment overnight. He has to write n lines of code before morning. He is dead tired and he tries drinking some black coffee to keep him awake. But each time he drinks a cup of coffee he stays awake for a short amount of time but his productivity goes down by a constant factor k

This is how he plans to write the program. He will write the first v lines of code, then drink his first cup of coffee. Since his productivity has gone down by a factor of k he will write v // k lines of code. He will have another cup of coffee and then write v // k**2 lines of code. He will have another cup of coffee and write v // k**3 lines of code and so on. He will collapse and fall asleep when v // k ** p becomes 0.

Now Vyasa does want to complete his assignment and maximize on his sleep. So he wants to figure out the minimum allowable value of v for a given productivity factor that will allow him to write at least n lines of code before he falls asleep.

Your input file will be called work.txt. Here is a typical file:

2 300 2 59 9 

The first line is T the number of test cases. This will be followed by T lines of input. Each line of input will have two numbers n and k. n is the number of lines of code to write and k is the productivity factor, where 1 n 10⁶ and 2 k 10.

For each test case your output to the screen will be v lines of code the Vyasa has to write, as well as the time it took for each function. For the above two test cases, the output will be:

Binary Search: 152 Time: 9.512901306152344e-05 Linear Search: 152 Time: 0.0005910396575927734 Binary Search: 54 Time: 4.696846008300781e-05 Linear Search: 54 Time: 9.012222290039062e-05 

Do not worry if your times don't match exactly. For this assignment, main has been written completely for you, and nothing needs to be changed in it.

You will be solving this problem in 2 ways. First, you will write a function that uses a linear search to solve the problem. Then you will write a function that uses a modified binary search algorithm to solve it again. Both functions will return the same answer, but the binary search method will usually be faster.

It is recommended that you write a helper function, which given a value v representing the number of lines Vyasa writes before his first cup of coffee and a value k, the productivity factor, will calculate the number of lines Vyasa will write before falling asleep. This can be called in both the linear and binary functions to make the computations easier.

The work.txt file contains:

2 300 2 59 9

Please use the template provided below

import time # Input: int n, the number of lines of code to write # int k, the productivity factor # Output: the number of lines of code that must be # written before the first cup of coffee def linear_search(n: int, k: int) -> int: # use linear search here return 0 # placeholder # Input: int n, the number of lines of code to write # int k, the productivity factor # Output: the number of lines of code that must be # written before the first cup of coffee def binary_search (n: int, k: int) -> int: # use binary search here return 0 # placeholder # main has been completed for you # do NOT change anything below this line def main(): in_file = open("work.txt", "r") num_cases = int((in_file.readline()).strip()) for i in range(num_cases): inp = (in_file.readline()).split() n = int(inp[0]) k = int(inp[1]) start = time.time() print("Binary Search: " + str(binary_search(n, k))) finish = time.time() print("Time: " + str(finish - start)) print() start = time.time() print("Linear Search: " + str(linear_search(n, k))) finish = time.time() print("Time: " + str(finish - start)) print() print() # The line above main is for grading purposes only. # DO NOT REMOVE THE LINE ABOVE MAIN if __name__ == "__main__": main()