code in c

int main() {

int n;
scanf("%d",&n);
unsigned long long int sum=0;
 for (long i =2; i<1000000; i++) {
     long temp = i;
     unsigned long long int sum1 = 0;
     while (temp>0) {
         int a = temp%10;
         sum1 +=(unsigned long long int) pow(a, n);
         temp = temp/10;
     }
     if (sum1 == i) {
           sum += sum1;
     }
 }
 printf("%lld",sum);
return 0;

}

8 months ago+ 0 comments

C++

#include <bits/stdc++.h>
using namespace std;

int main() {
     
    int n;
    cin >> n;
    vector<int> arr(10);
    for (int i=0; i<=9; i++)
    {
        int p = pow(i, n);
        arr[i] = p;
    }
    int S = 0;
    for (int i=100; i<1000000; i++)
    {
        int num = i, sum=0;
        while(num>0)
        {
            int z = num%10;
            sum = sum+arr[z];
            num = num/10;
        }
        if (sum==i) S = S + sum;
    }
    cout << S;
    return 0;
}

2 years ago+ 1 comment

My sense was to do the following.

MIN = 2
MAX = 1000000


def compute(value):
    ans = sum(i for i in range(MIN, MAX)
              if i == power_digit_sum(value, i))
    return str(ans)


def power_digit_sum(value, n):
    return sum(int(c) ** value for c in str(n))


if __name__ == "__main__":
    value = int(input())
    print(compute(value))

2 years ago+ 1 comment

My Approach - First find all possible numbers that a K-digit number could have for example I treated [1,2,3],[2,1,3],[1,3,2],[2,3,1],[3,1,2],[3,2,1] all as same and I find only one of them and not repeated numbers with same digits (using recursive appraoch) since what we are interested in only digits which constitute K-digit number . this way my algorithm is very fast and will also work if N (input) is even more than 6 .

3 years ago+ 0 comments

You can still use brute-force. Just do some experiments to determine lower and upper limit for i, then you will pass all cases.

def Digit_Nth_Powers(n,lower,upper):
  sumList = 0

  for i in range(lower,upper):
    digits = list(map(int,str(i)))
    tempSum = 0

    for j in digits:
      tempSum += j**n
    
    if i == tempSum:
      sumList += i
        
  return sumList

if __name__ == '__main__':
  n = int(input())
  
  # Lower and upper limit is determined by experiments for faster code execution
  if n == 3:
    lower,upper = [1*(10**2), 1*(10**3)]
  elif n == 4:
    lower,upper = [1*(10**3), 1*(10**4)]  
  elif n == 5:
    lower,upper = [2*(10**3), 2*(10**5)]
  elif n == 6:
    lower,upper = [1*(10**5), 6*(10**5)]
    
  sumList = Digit_Nth_Powers(n,lower,upper)
  print(sumList)

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