Nepietiekams numurs
Izmēģiniet to GfG Practice 
#practiceLinkDiv { display: none !important; }
#practiceLinkDiv { display: none !important; } Tiek uzskatīts, ka skaitlis n ir nepilnīgs skaitlis, ja tā ir visu skaitļa dalītāju summa, kas apzīmēta ar dalītājiSumma(n) ir mazāks par divkāršu skaitļa n vērtību. Un atšķirību starp šīm divām vērtībām sauc par trūkums .
Matemātiski, ja zemāk esošais nosacījums atbilst, skaitlis tiek uzskatīts par nepilnīgu:
divisorsSum(n) < 2 * n deficiency = (2 * n) - divisorsSum(n)
Pirmie nepilnīgie skaitļi ir:
1 2 3 4 5 7 8 9 10 11 13 14 15 16 17 19 .....
Dots skaitlis n, mūsu uzdevums ir noskaidrot, vai šis skaitlis ir vai nav.
Piemēri:
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
A Vienkāršs risinājums ir atkārtot visus skaitļus no 1 līdz n un pārbaudīt, vai skaitlis dala n, un aprēķināt summu. Pārbaudiet, vai šī summa ir mazāka par 2 * n.
Laiks Šīs pieejas sarežģītība: O ( n )
Optimizēts risinājums:
Ja uzmanīgi vērojam, skaitļa n dalītāji ir pa pāriem. Piemēram, ja n = 100, tad visi dalītāju pāri ir: (1 100) (2 50) (4 25) (5 20) (10 10)
Izmantojot šo faktu, mēs varam paātrināt mūsu programmu.
Pārbaudot dalītājus, mums būs jābūt uzmanīgiem, ja ir divi vienādi dalītāji, kā gadījumā (10 10). Šādā gadījumā summas aprēķinā ņemsim tikai vienu no tiem.
Optimizētās pieejas ieviešana
// C++ program to implement an Optimized Solution // to check Deficient Number #include using namespace std ; // Function to calculate sum of divisors 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 <= 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 bool isDeficient ( int n ) { // Check if sum(n) < 2 * n return ( divisorsSum ( n ) < ( 2 * n )); } /* Driver program to test above function */ int main () { isDeficient ( 12 ) ? cout < < 'YES n ' : cout < < 'NO n ' ; isDeficient ( 15 ) ? cout < < 'YES n ' : cout < < 'NO n ' ; return 0 ; }
Java // 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
Python3 # 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# // 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 // 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 ) ? 'YES n ' : 'NO n ' ; echo ( $ds ); $ds = isDeficient ( 15 ) ? 'YES n ' : 'NO n ' ; echo ( $ds ); // This code is contributed by ajit;. ?>
JavaScript < 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>
Izvade:
NO YES
Laika sarežģītība: O(sqrt(n))
Palīgtelpa: O(1)
Atsauces:
https://en.wikipedia.org/wiki/Deficient_number
