#practiceLinkDiv { display: none !important; }Dots naturāls skaitlis n uzdevums ir atrast visu n dalītāju dalītāju summu.
pamēģini noķert java
Piemēri:
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 Atrodiet dalītāju summu Izmēģiniet to!
Izmantojot to, ka jebkurš skaitlis n var izteikt kā galveno faktoru reizinājumu n = lpp1k1x lpp2k2x ... kur p1lpp2... ir pirmskaitļi.
Visus n dalītājus var izteikt kā p1ax lpp2bx ... kur 0<= a <= k1 and 0 <= b <= k2.
Tagad dalītāju summa būs p visu pakāpju summa1- lpp1lpp11.... lpp1k1reizināts ar visu p2- lpp2lpp21.... lpp2k1
n dalītāja summa
= (lpp1x lpp2) + (lpp11x lpp2) +.....+ (lpp1k1x lpp2) +...+ (lpp1x lpp21) + (lpp11x lpp21) +.....+ (lpp1k1x lpp21) +........+
(lpp1x lpp2k2) + (lpp11x lpp2k2) +......+ (lpp1k1x lpp2k2).
= (lpp1+ lpp11+...+ lpp1k1) x p2+ (lpp1+ lpp11+...+ lpp1k1) x p21+.......+ (lpp1+ lpp11+...+ lpp1k1) x p2k2.
= (lpp1+ lpp11+...+ lpp1k1) x (lpp2+ lpp21+...+ lpp2k2).
Tagad dalītāji jebkura pap kā pirmskaitļi ir plpp1...... lppa. Un dalītāju summa būs (lpp(a+1)- 1)/(p -1) ļauj to definēt ar f(p).
Tātad visu dalītāju dalītāju summa būs
= (f(p1) + f(lpp11) +...+ f(lpp1k1)) x (f(p2) + f(lpp21) +...+ f(lpp2k2)).
Tātad, ņemot vērā skaitli n ar primāro faktorizāciju, mēs varam atrast visu dalītāju dalītāju summu. Bet šajā uzdevumā mums ir dots, ka n ir masīva elementa reizinājums. Tātad atrodiet katra elementa primāro faktorizāciju un, izmantojot faktu abx ac= ab+c.
Tālāk ir sniegta informācija par šīs pieejas īstenošanu.
C++// C++ program to find sum of divisors of all // the divisors of a natural number. #include
// 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 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# 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
<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>
Izvade:
28
Laika sarežģītība: O(?n log n)
Palīgtelpa: O(n)
Optimizācijas:
Gadījumos, kad ir vairākas ievades, kurām mums ir jāatrod vērtība, ko varam izmantot Eratostena siets kā apspriests šis pastu.