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72 lines
2.1 KiB
Java
72 lines
2.1 KiB
Java
package com.thealgorithms.maths;
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/**
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* In number theory, a perfect number is a positive integer that is equal to the
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* sum of its positive divisors, excluding the number itself. For instance, 6
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* has divisors 1, 2 and 3 (excluding itself), and 1 + 2 + 3 = 6, so 6 is a
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* perfect number.
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*
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* link:https://en.wikipedia.org/wiki/Perfect_number
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*/
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public final class PerfectNumber {
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private PerfectNumber() {
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}
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/**
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* Check if {@code number} is perfect number or not
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*
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* @param number the number
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* @return {@code true} if {@code number} is perfect number, otherwise false
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*/
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public static boolean isPerfectNumber(int number) {
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if (number <= 0) {
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return false;
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}
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int sum = 0;
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/* sum of its positive divisors */
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for (int i = 1; i < number; ++i) {
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if (number % i == 0) {
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sum += i;
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}
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}
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return sum == number;
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}
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/**
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* Check if {@code n} is perfect number or not
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*
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* @param n the number
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* @return {@code true} if {@code number} is perfect number, otherwise false
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*/
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public static boolean isPerfectNumber2(int n) {
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if (n <= 0) {
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return false;
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}
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int sum = 1;
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double root = Math.sqrt(n);
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/*
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* We can get the factors after the root by dividing number by its factors
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* before the root.
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* Ex- Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50 and 100.
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* Root of 100 is 10. So factors before 10 are 1, 2, 4 and 5.
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* Now by dividing 100 by each factor before 10 we get:
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* 100/1 = 100, 100/2 = 50, 100/4 = 25 and 100/5 = 20
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* So we get 100, 50, 25 and 20 which are factors of 100 after 10
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*/
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for (int i = 2; i <= root; i++) {
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if (n % i == 0) {
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sum += i + n / i;
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}
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}
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// if n is a perfect square then its root was added twice in above loop, so subtracting root
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// from sum
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if (root == (int) root) {
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sum -= (int) root;
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}
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return sum == n;
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}
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}
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