My explanation of the recursive approach. Since there isn't an editorial.

What we have: num and pow (N and power)

Before we even solve the problem we have to solve another problem: What's the max integer value, such that value^pow<=num

int max_value=(int) num^(1/pow) #max=(num**(1.to_f/$pow)).to_i this is my line in the code

Why do we care about this max_value? because the set (1..max_value) i.e. from 1 to max_value inclusive are the only values s.t. values^pow <= num, any number greater than this range is greater than the num and not worth computing time...

So we have a range: 1 to max_value. This will help with the base case.

Now the recursion, is a little more complex than just one call...it's two calls. lol. It honestly, took me too long to figure this part out.

num=gets().to_i
$pow=gets().to_i
max=(num**(1.to_f/$pow)).to_i
def recurse(n,max)
    return 0 if (n < 0 || (n > 0 && max == 0))
    return 1 if (n == 0)
    return recurse(n-(max**$pow),max-1)+recurse(n,max-1)
end
puts recurse(num,max)

Recursive Call:

Don't worry yet: recursive(num, pow, max) is the first call we'll make.

we'll return the sum of two things:

recursive(num-(max^pow), pow, max-1)

num-max^pow is the new value we'll search. max-1 exists because we want to take max out of circulation and only count values s.t.(1 to max-1)

The other part is

recursive(num,pow,max-1)

Same reasoning as above, but you only want solutions that do NOT include max.

Explaining the recursive Base cases:

if num<0, then we subtracted too much and no value exists.

if num>0 and max==0, then there is a remainder in the subtraction, and no remaining numbers to subtract

if num==0, then we have 1 instance where the sum is possible.