Note: The input number n will not exceed 100,000,000. (1e8)
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publicclassSolution{ publicbooleancheckPerfectNumber(int num){ if (num < 2) returnfalse; int sqrt = (int) Math.sqrt(num); int result = 1; for (int i = 2; i <= sqrt; i++) { if (num % i == 0) { result += i + (num / i); } } if (sqrt * sqrt == num) { result -= sqrt; }