Here is Python code.

Time complexity O(n)

# import inbuilt standard input output
from sys import stdin, stdout

# suppose a function called main() and
# all the operations are performed
def main():

    # input via readline method
    n = stdin.readline()

    # array input similar method
    arr = [int(i) for i in stdin.readline().split()]

    arr.sort()

    #initialize variable
    t = int(n)-1
    p = float(1/t)
    
    temp = int((int(arr[t])**p )//1)
    ans = int(0)
    ans1 = int(0)
    
    for i in range(int(n)):
        ans = int(int(ans) + int(abs(int(int(temp**i) - arr[i]))))
    
    temp = int(int(temp) + 1)
    
    for i in range(int(n)):
        ans1 = int(int(ans1) + int(abs(int(int(temp**i) - arr[i]))))
    
    ans = min(int(ans), int(ans1))
    
    stdout.write(str(ans))

# call the main method
if __name__ == "__main__":
    main()

I try another approach in Problem B.

First sort the array

then find the max element called it "max", which is last element of array, because we sorted the array.

Now we take (n-1) root of "max" called it "r"

here "r" may be not integer, so we have to check for two numbers a=floor(r) and b=ceil(r) eg. if r = 5.67 so we take a = 5 and b = 6

now we generate two new power array from this two number and sum of abs different between all element of this array and our sorted array do this for both "a" and "b"

and ans will me minimum of this two.