Problema
Add the positive divisors of n that are strictly smaller than n. If their sum equals n, the number is perfect. If the sum is larger, it is abundant; if smaller, it is deficient. The number 1 has no proper positive divisors, so its divisor sum is 0.
Implement perfect_number(n).
Input
n: The positive integer to classify.
Output
Return perfect, abundant, or deficient as lowercase text.
Vincoli
1 ≤ n ≤ 1,000,000. Zero and negative values are excluded.
Esempi
Input:
n = 18
Output atteso:
expected = "abundant"Input:
long long n = 18;
Output atteso:
std::string expected = "abundant";Input:
long n = 18;
Output atteso:
string expected = "abundant";Spiegazione
The proper divisors are 1, 2, 3, 6, and 9. Their sum is 21, greater than 18.
Input:
n = 28
Output atteso:
expected = "perfect"Input:
long long n = 28;
Output atteso:
std::string expected = "perfect";Input:
long n = 28;
Output atteso:
string expected = "perfect";Spiegazione
The proper divisors sum to 1 + 2 + 4 + 7 + 14 = 28, exactly the number itself.
Come affrontarlo
Prova prima da solo, poi apri un suggerimento alla volta.
Suggerimento 1
Find the proper divisors of n and compare their sum with n to decide the class.
Suggerimento 2
You only need to search divisors up to the square root, pairing each divisor with its companion.
Suggerimento 3
Treat 1 carefully, since it has no proper positive divisors and its divisor sum is zero.
Complessità attesa
O(√n) time, O(1) extra space
Riferimenti
Questa sfida è adattata per CodeUtility. Segui la traccia e i vincoli sopra.