Calcular la suma de los divisores de todos los divisores de un número natural.
#practiceLinkDiv { mostrar: ninguno !importante; } Dado un numero natural norte la tarea es encontrar la suma de los divisores de todos los divisores de n.
Ejemplos:
Input : n = 54 Output : 232 Divisors of 54 = 1 2 3 6 9 18 27 54. Sum of divisors of 1 2 3 6 9 18 27 54 are 1 3 4 12 13 39 40 120 respectively. Sum of divisors of all the divisors of 54 = 1 + 3 + 4 + 12 + 13 + 39 + 40 + 120 = 232. Input : n = 10 Output : 28 Divisors of 10 are 1 2 5 10 Sums of divisors of divisors are 1 3 6 18. Overall sum = 1 + 3 + 6 + 18 = 28Recommended Practice encontrar la suma de divisores ¡Pruébalo!
Usando el hecho de que cualquier número norte se puede expresar como producto de factores primos norte =p 1 k1 xp 2 k2 x... donde p 1 pag 2 ... son números primos.
Todos los divisores de n se pueden expresar como p. 1 a xp 2 b x... donde 0 <= a <= k1 and 0 <= b <= k2.
Ahora la suma de los divisores será la suma de todas las potencias de p. 1 - pag 1 pag 1 1 .... pag 1 k1 multiplicado por todo el poder de p 2 - pag 2 pag 2 1 .... pag 2 k1
Suma del divisor de n
= (pag 1 xp 2 ) + (pag 1 1 xp 2 ) +.....+ (p. 1 k1 xp 2 ) +....+ (p. 1 xp 2 1 ) + (pag 1 1 xp 2 1 ) +.....+ (p. 1 k1 xp 2 1 ) +.......+
(pag 1 xp 2 k2 ) + (pag 1 1 xp 2 k2 ) +......+ (p. 1 k1 xp 2 k2 ).
= (pag 1 +p 1 1 +...+ p 1 k1 ) xp 2 + (pag. 1 +p 1 1 +...+ p 1 k1 ) xp 2 1 +.......+ (p. 1 +p 1 1 +...+ p 1 k1 ) xp 2 k2 .
= (pag 1 +p 1 1 +...+ p 1 k1 ) x (pag 2 +p 2 1 +...+ p 2 k2 ).
Ahora los divisores de cualquier p a para p como primo son p pag 1 ...... pag a . Y la suma de divisores será (p (un+1) - 1)/(p -1) déjelo definir por f(p).
Entonces la suma de los divisores de todos los divisores será
= (f(p 1 ) + f(p 1 1 ) +...+ f(pag 1 k1 )) x (f(pag 2 ) + f(p 2 1 ) +...+ f(pag 2 k2 )).
Entonces, dado un número n, mediante factorización prima podemos encontrar la suma de los divisores de todos los divisores. Pero en este problema se nos da que n es producto del elemento de la matriz. Entonces encuentre la factorización prima de cada elemento y usando el hecho de a b x un do = un b+c .
A continuación se muestra la implementación de este enfoque:
C++ // C++ program to find sum of divisors of all // the divisors of a natural number. #include using namespace std ; // Returns sum of divisors of all the divisors // of n int sumDivisorsOfDivisors ( int n ) { // Calculating powers of prime factors and // storing them in a map mp[]. map < int int > mp ; for ( int j = 2 ; j <= sqrt ( n ); j ++ ) { int count = 0 ; while ( n % j == 0 ) { n /= j ; count ++ ; } if ( count ) mp [ j ] = count ; } // If n is a prime number if ( n != 1 ) mp [ n ] = 1 ; // For each prime factor calculating (p^(a+1)-1)/(p-1) // and adding it to answer. int ans = 1 ; for ( auto it : mp ) { int pw = 1 ; int sum = 0 ; for ( int i = it . second + 1 ; i >= 1 ; i -- ) { sum += ( i * pw ); pw *= it . first ; } ans *= sum ; } return ans ; } // Driven Program int main () { int n = 10 ; cout < < sumDivisorsOfDivisors ( n ); return 0 ; }
Java // Java program to find sum of divisors of all // the divisors of a natural number. import java.util.HashMap ; class GFG { // Returns sum of divisors of all the divisors // of n public static int sumDivisorsOfDivisors ( int n ) { // Calculating powers of prime factors and // storing them in a map mp[]. HashMap < Integer Integer > mp = new HashMap <> (); for ( int j = 2 ; j <= Math . sqrt ( n ); j ++ ) { int count = 0 ; while ( n % j == 0 ) { n /= j ; count ++ ; } if ( count != 0 ) mp . put ( j count ); } // If n is a prime number if ( n != 1 ) mp . put ( n 1 ); // For each prime factor calculating (p^(a+1)-1)/(p-1) // and adding it to answer. int ans = 1 ; for ( HashMap . Entry < Integer Integer > entry : mp . entrySet ()) { int pw = 1 ; int sum = 0 ; for ( int i = entry . getValue () + 1 ; i >= 1 ; i -- ) { sum += ( i * pw ); pw *= entry . getKey (); } ans *= sum ; } return ans ; } // Driver code public static void main ( String [] args ) { int n = 10 ; System . out . println ( sumDivisorsOfDivisors ( n )); } } // This code is contributed by // sanjeev2552
Python3 # Python3 program to find sum of divisors # of all the divisors of a natural number. import math as mt # Returns sum of divisors of all # the divisors of n def sumDivisorsOfDivisors ( n ): # Calculating powers of prime factors # and storing them in a map mp[]. mp = dict () for j in range ( 2 mt . ceil ( mt . sqrt ( n ))): count = 0 while ( n % j == 0 ): n //= j count += 1 if ( count ): mp [ j ] = count # If n is a prime number if ( n != 1 ): mp [ n ] = 1 # For each prime factor calculating # (p^(a+1)-1)/(p-1) and adding it to answer. ans = 1 for it in mp : pw = 1 summ = 0 for i in range ( mp [ it ] + 1 0 - 1 ): summ += ( i * pw ) pw *= it ans *= summ return ans # Driver Code n = 10 print ( sumDivisorsOfDivisors ( n )) # This code is contributed # by mohit kumar 29
C# // C# program to find sum of divisors of all // the divisors of a natural number. using System ; using System.Collections.Generic ; class GFG { // Returns sum of divisors of // all the divisors of n public static int sumDivisorsOfDivisors ( int n ) { // Calculating powers of prime factors and // storing them in a map mp[]. Dictionary < int int > mp = new Dictionary < int int > (); for ( int j = 2 ; j <= Math . Sqrt ( n ); j ++ ) { int count = 0 ; while ( n % j == 0 ) { n /= j ; count ++ ; } if ( count != 0 ) mp . Add ( j count ); } // If n is a prime number if ( n != 1 ) mp . Add ( n 1 ); // For each prime factor // calculating (p^(a+1)-1)/(p-1) // and adding it to answer. int ans = 1 ; foreach ( KeyValuePair < int int > entry in mp ) { int pw = 1 ; int sum = 0 ; for ( int i = entry . Value + 1 ; i >= 1 ; i -- ) { sum += ( i * pw ); pw = entry . Key ; } ans *= sum ; } return ans ; } // Driver code public static void Main ( String [] args ) { int n = 10 ; Console . WriteLine ( sumDivisorsOfDivisors ( n )); } } // This code is contributed // by Princi Singh
JavaScript < script > // Javascript program to find sum of divisors of all // the divisors of a natural number. // Returns sum of divisors of all the divisors // of n function sumDivisorsOfDivisors ( n ) { // Calculating powers of prime factors and // storing them in a map mp[]. let mp = new Map (); for ( let j = 2 ; j <= Math . sqrt ( n ); j ++ ) { let count = 0 ; while ( n % j == 0 ) { n = Math . floor ( n / j ); count ++ ; } if ( count != 0 ) mp . set ( j count ); } // If n is a prime number if ( n != 1 ) mp . set ( n 1 ); // For each prime factor calculating (p^(a+1)-1)/(p-1) // and adding it to answer. let ans = 1 ; for ( let [ key value ] of mp . entries ()) { let pw = 1 ; let sum = 0 ; for ( let i = value + 1 ; i >= 1 ; i -- ) { sum += ( i * pw ); pw = key ; } ans *= sum ; } return ans ; } // Driver code let n = 10 ; document . write ( sumDivisorsOfDivisors ( n )); // This code is contributed by patel2127 < /script>
Producción:
28
Complejidad del tiempo: O (? norte iniciar sesión norte)
Espacio Auxiliar: En)
Optimizaciones:
Para los casos en los que hay varias entradas para las que necesitamos encontrar el valor que podemos usar Tamiz de Eratóstenes como se discutió en este correo.
