Problème
Number the digits of n from the left, starting at 1. Raise each digit to the power of its position and add the results. For 352 that is 3¹ + 5² + 2³. A zero digit adds 0.
Implement digit_power_sum(n).
Input
n: A positive integer.
Output
Return the sum of the powered digits.
Contraintes
1 ≤ n ≤ 1,000,000,000
Exemples
Exemple 1
Entrée :
n = 352
Sortie attendue :
expected = 36Entrée :
long long n = 352;
Sortie attendue :
long long expected = 36;Entrée :
long n = 352;
Sortie attendue :
long expected = 36;Explication
3¹ + 5² + 2³ = 3 + 25 + 8 = 36.
Exemple 2
Entrée :
n = 89
Sortie attendue :
expected = 89Entrée :
long long n = 89;
Sortie attendue :
long long expected = 89;Entrée :
long n = 89;
Sortie attendue :
long expected = 89;Explication
8¹ + 9² = 8 + 81 = 89, the number itself.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
Scan the number digit by digit from left to right and keep track of each digit’s position.
Indice 2
Convert the number to a string or otherwise know the left-to-right index before combining each digit with its position.
Indice 3
Remember that a zero digit contributes nothing, but its position still matters for the exponent count.
Complexité attendue
O(d) time, O(1) extra space