#practiceLinkDiv { display: none !important; }Za broj n se kaže da je manjkav broj ako je zbroj svih djelitelja broja označen s djeliteljiZbroj(n) manja je od dvostruke vrijednosti broja n. A razlika između ove dvije vrijednosti naziva se nedostatak .
Matematički gledano, ako vrijedi donji uvjet, broj se smatra manjkavim:
divisorsSum(n) < 2 * n deficiency = (2 * n) - divisorsSum(n)
Prvih nekoliko manjkavih brojeva su:
1 2 3 4 5 7 8 9 10 11 13 14 15 16 17 19 .....
Zadan je broj n, naš zadatak je pronaći je li taj broj Manjavi broj ili nije.
Primjeri:
Input: 21 Output: YES Divisors are 1 3 7 and 21. Sum of divisors is 32. This sum is less than 2*21 or 42. Input: 12 Output: NO Input: 17 Output: YES
stog u dsPreporučena praksa Nedostatak broja Probajte!
A Jednostavno rješenje je ponoviti sve brojeve od 1 do n i provjeriti dijeli li broj n te izračunati zbroj. Provjerite je li taj zbroj manji od 2 * n ili nije.
Vremenska složenost ovog pristupa: O ( n )
Optimizirano rješenje:
Ako pažljivo promatramo, djelitelji broja n prisutni su u parovima. Na primjer, ako je n = 100, tada su svi parovi djelitelja: (1 100) (2 50) (4 25) (5 20) (10 10)
Koristeći ovu činjenicu možemo ubrzati naš program.
Prilikom provjere djelitelja morat ćemo paziti ako postoje dva jednaka djelitelja kao u slučaju (10 10). U tom slučaju ćemo uzeti samo jedan od njih u izračun zbroja.
Implementacija optimiziranog pristupa
// C++ program to implement an Optimized Solution // to check Deficient Number #include
// Java program to check Deficient Number import java.io.*; class GFG { // Function to calculate sum of divisors static int divisorsSum(int n) { int sum = 0; // Initialize sum of prime factors // Note that this loop runs till square root of n for (int i = 1; i <= (Math.sqrt(n)); i++) { if (n % i == 0) { // If divisors are equal take only one // of them if (n / i == i) { sum = sum + i; } else // Otherwise take both { sum = sum + i; sum = sum + (n / i); } } } return sum; } // Function to check Deficient Number static boolean isDeficient(int n) { // Check if sum(n) < 2 * n return (divisorsSum(n) < (2 * n)); } /* Driver program to test above function */ public static void main(String args[]) { if (isDeficient(12)) System.out.println('YES'); else System.out.println('NO'); if (isDeficient(15)) System.out.println('YES'); else System.out.println('NO'); } } // This code is contributed by Nikita Tiwari
# Python program to implement an Optimized # Solution to check Deficient Number import math # Function to calculate sum of divisors def divisorsSum(n) : sum = 0 # Initialize sum of prime factors # Note that this loop runs till square # root of n i = 1 while i<= math.sqrt(n) : if (n % i == 0) : # If divisors are equal take only one # of them if (n // i == i) : sum = sum + i else : # Otherwise take both sum = sum + i; sum = sum + (n // i) i = i + 1 return sum # Function to check Deficient Number def isDeficient(n) : # Check if sum(n) < 2 * n return (divisorsSum(n) < (2 * n)) # Driver program to test above function if ( isDeficient(12) ): print ('YES') else : print ('NO') if ( isDeficient(15) ) : print ('YES') else : print ('NO') # This Code is contributed by Nikita Tiwari
// C# program to implement an Optimized Solution // to check Deficient Number using System; class GFG { // Function to calculate sum of // divisors static int divisorsSum(int n) { // Initialize sum of prime factors int sum = 0; // Note that this loop runs till // square root of n for (int i = 1; i <= (Math.Sqrt(n)); i++) { if (n % i == 0) { // If divisors are equal // take only one of them if (n / i == i) { sum = sum + i; } else // Otherwise take both { sum = sum + i; sum = sum + (n / i); } } } return sum; } // Function to check Deficient Number static bool isDeficient(int n) { // Check if sum(n) < 2 * n return (divisorsSum(n) < (2 * n)); } /* Driver program to test above function */ public static void Main() { string var = isDeficient(12) ? 'YES' : 'NO'; Console.WriteLine(var); string var1 = isDeficient(15) ? 'YES' : 'NO'; Console.WriteLine(var1); } } // This code is contributed by vt_m
// PHP program to implement // an Optimized Solution // to check Deficient Number // Function to calculate // sum of divisors function divisorsSum($n) { // Initialize sum of // prime factors $sum = 0; // Note that this loop runs // till square root of n for ($i = 1; $i <= sqrt($n); $i++) { if ($n % $i==0) { // If divisors are equal // take only one of them if ($n / $i == $i) { $sum = $sum + $i; } // Otherwise take both else { $sum = $sum + $i; $sum = $sum + ($n / $i); } } } return $sum; } // Function to check // Deficient Number function isDeficient($n) { // Check if sum(n) < 2 * n return (divisorsSum($n) < (2 * $n)); } // Driver Code $ds = isDeficient(12) ? 'YESn' : 'NOn'; echo($ds); $ds = isDeficient(15) ? 'YESn' : 'NOn'; echo($ds); // This code is contributed by ajit;. ?> <script> // Javascript program to check Deficient Number // Function to calculate sum of divisors function divisorsSum(n) { let sum = 0; // Initialize sum of prime factors // Note that this loop runs till square root of n for (let i = 1; i <= (Math.sqrt(n)); i++) { if (n % i == 0) { // If divisors are equal take only one // of them if (n / i == i) { sum = sum + i; } else // Otherwise take both { sum = sum + i; sum = sum + (n / i); } } } return sum; } // Function to check Deficient Number function isDeficient(n) { // Check if sum(n) < 2 * n return (divisorsSum(n) < (2 * n)); } // Driver code to test above methods if (isDeficient(12)) document.write('YES' + '
'); else document.write('NO' + '
'); if (isDeficient(15)) document.write('YES' + '
'); else document.write('NO' + '
'); // This code is contributed by avijitmondal1998. </script>
Izlaz:
NO YES
Vremenska složenost: O(sqrt(n))
Pomoćni prostor: O(1)
Reference:
https://en.wikipedia.org/wiki/Deficient_number