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.
Restrições
1 ≤ n ≤ 1,000,000. Zero and negative values are excluded.
Exemplos
Entrada:
n = 18
Saída esperada:
expected = "abundant"Entrada:
long long n = 18;
Saída esperada:
std::string expected = "abundant";Entrada:
long n = 18;
Saída esperada:
string expected = "abundant";Explicação
The proper divisors are 1, 2, 3, 6, and 9. Their sum is 21, greater than 18.
Entrada:
n = 28
Saída esperada:
expected = "perfect"Entrada:
long long n = 28;
Saída esperada:
std::string expected = "perfect";Entrada:
long n = 28;
Saída esperada:
string expected = "perfect";Explicação
The proper divisors sum to 1 + 2 + 4 + 7 + 14 = 28, exactly the number itself.
Como abordar
Tente sozinho primeiro e depois abra uma dica de cada vez.
Dica 1
Find the proper divisors of n and compare their sum with n to decide the class.
Dica 2
You only need to search divisors up to the square root, pairing each divisor with its companion.
Dica 3
Treat 1 carefully, since it has no proper positive divisors and its divisor sum is zero.
Complexidade esperada
O(√n) time, O(1) extra space
Referências
Este desafio foi adaptado para o CodeUtility. Siga o enunciado e as restrições acima.