We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Recursion
  4. The Power Sum
  5. Discussions

The Power Sum

Sort by

recency

|

353 Discussions

|

  • weaslypaul923
     + 0 comments

    I was experimenting with different ways to season my meals and stumbled upon the unique taste of Himalayan salt, which got me curious about connecting with a reliable Himalayan salt exporter. Using it brought out flavors in dishes I never expected, and I realized that working with a Himalayan salt exporter could make it easier to have high-quality salt at home consistently. It felt satisfying knowing exactly where my ingredients were coming from and how pure they were. Sharing a pinch with friends made cooking feel more intentional, and having access through a Himalayan salt exporter makes it simple to keep experimenting. The whole experience turned a basic meal into something special without much effort.

  • vutrantrung14821
     + 0 comments
    def h(i,S,X,N):
    e=0
    for j in range(i+1,int(X**1/N)+1):
        if S+j**N<X:
            e=e+h(j,S+j**N,X,N)
        elif S+j**N==X:
            e+=1
        else:
            continue
    return e



def powerSum(X, N):
    a=0
    for i in range(1,int(X**1/N)+1):
        S=i**N
        if S < X:
            a += h(i, S, X, N)
        elif S == X:
            a += 1
    return a

  • ilovemylittlecr1
     + 1 comment

    The most intuitive solution you will ever find in this entire discussion. Plus, the run time speed surpassed editorial solution.

    include

    using namespace std;

    int ipow(const int& b, const int& e){ if(e == 0) return 1; return b * pow(b, e - 1); }

    int recursion_ans(const vector& vec, const int& index, const int& remaining){ if(remaining == 0) return 1; if(index < 0 || remaining < 0) return 0;

    return recursion_ans(vec, index - 1, remaining - vec[index]) + 
        recursion_ans(vec, index - 1, remaining);

    }

    int main(){ int X, N; cin >> X >> N;

    vector<int> vec;

int i = 1;
while(i){
    int value = ipow(i, N); ++i;
    if(value > X) break;
    vec.push_back(value);
}

cout << recursion_ans(vec, vec.size() - 1, X);

return 0;

    }

  • jmcparland
     + 0 comments

    Haskell

    Function wwo -- "with or without"

    module Main where

vals :: Int -> Int -> [Int]
vals n k = reverse $ takeWhile (<= n) [x ^ k | x <- [1 ..]]

wwo :: Int -> [Int] -> Int
wwo n xs
  | n == 0 = 1
  | n < 0 = 0
  | null xs = 0
  | otherwise = wwo n (tail xs) + wwo (n - head xs) (tail xs)

main :: IO ()
main = do
  n <- readLn :: IO Int
  k <- readLn :: IO Int
  print $ wwo n (vals n k)

  • SI006677Abhishek
     + 1 comment

    int findPows(int X, int num, int N) {

    int val = X - pow(num, N);
if (val < 0) {
    return 0;  
}
if (val == 0) {
    return 1;  
}

return findPows(val, num + 1, N) + findPows(X, num + 1, N);

    }

    int powerSum(int X, int N) { return findPows(X, 1, N); }