#practiceLinkDiv { display: ingen !important; }Et tal n siges at være mangelfuldt tal, hvis summen af alle divisorerne af tallet angivet med divisorsSum(n) er mindre end to gange værdien af tallet n. Og forskellen mellem disse to værdier kaldes mangel .
Matematisk, hvis nedenstående betingelse gælder, siges tallet at være mangelfuldt:
divisorsSum(n) < 2 * n deficiency = (2 * n) - divisorsSum(n)
De første par mangelfulde tal er:
1 2 3 4 5 7 8 9 10 11 13 14 15 16 17 19 .....
Givet et tal n er vores opgave at finde ud af, om dette tal er mangelfuldt tal eller ej.
Eksempler:
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
Anbefalet praksis Manglende nummer Prøv det!
EN Simpel løsning er at iterere alle tallene fra 1 til n og tjekke om tallet deler n og udregne summen. Tjek om denne sum er mindre end 2 * n eller ej.
Tidskompleksitet af denne tilgang: O ( n )
Optimeret løsning:
Hvis vi observerer nøje, er divisorerne for tallet n til stede i par. For eksempel hvis n = 100, så er alle divisorpar: (1 100) (2 50) (4 25) (5 20) (10 10)
Ved at bruge dette faktum kan vi fremskynde vores program.
Når vi tjekker divisorer, skal vi være forsigtige, hvis der er to lige store divisorer som i tilfælde af (10 10). I så fald tager vi kun én af dem i beregningen af summen.
Implementering af Optimeret tilgang
// 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>
Output:
NO YES
Tidskompleksitet: O( sqrt(n))
Hjælpeplads: O(1)
Referencer:
https://en.wikipedia.org/wiki/Deficient_number