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  3. Project Euler #56: Powerful digit sum
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Project Euler #56: Powerful digit sum

  • tharindukavinda1
     + 1 comment

    my solution in python

    # Enter your code here. Read input from STDIN. Print output to STDOUT

n= int (input())

#a , b =0
a=0
for i in range ( n ):
    for j in range (n):
        p=pow(i,j)
        #print (p)
        p=str(p)
        p=list(p)

        q=0

        for k in range (len(p)):
            q=q+ int (p[k])

        if ( a<q ):
            a=q


print(a)

    • mpbento
       + 1 comment

      There are quite a few things that can be improved for the performance obcessed:

      • a**b is much faster than pow(a, b). Try it with timeit.
      • You could iterate as: for k in p: q += int(k).
      • You could use tuple() instead of list(). Faster most of the times.

      Just a few thoughts.

      • vidush_rk11
         + 0 comments

        Are you sure that a* * b is faster than pow(a,b)?

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