# Sum of all proper divisors of natural numbers in an array

Given an array of natural numbers count the sum of its proper divisors for every element in array.

Input : int arr[] = {8, 13, 24, 36, 59, 75, 87} Output : 7 1 36 55 1 49 21 Number 8 has 3 proper divisors 1, 2, 4 and their sum comes out to be 7.

A naive solution to this problem has been discussed in below post.

Sum of all proper divisors of a natural number

We can do this more efficiently by making use of sieve of Eratosthenes.

The idea is based on prime factorization of a number. By using sieve we can store all the prime factors of a number and their powers.

To find all divisors, we need to consider all powers of a prime factor and multiply it with all all powers of other prime factors. (For example, if the number is 36, its prime factors are 2 and 3 and all divisors are 1, 2, 3, 4, 6, 9, 12 and 18. Consider a number N can be written as P1^Q1 * P2^Q2 * P3^Q3 (here only 3 prime factors are considered but there can be more than that) then sum of its divisors will be written as: = P1^0 * P2^0 * P3^0 + P1^0 * P2^0 * P3^1 + P1^0 * P2^0 * P3^2 + ................ + P1^0 * P2^0 * P3^Q3 + P1^0 * P2^1 * P3^0 + P1^0 * P2^1 * P3^1 + P1^0 * P2^1 * P3^2 + ................ + P1^0 * P2^1 * P3^Q3 + . . . P1^Q1 * P2^Q2 * P3^0 + P1^Q1 * P2^Q2 * P3^1 + P1^Q1 * P2^Q2 * P3^2 + .......... + P1^Q1 * P2^Q2 * P3^Q3 Above can be written as, (((P1^(Q1+1)) - 1) / (P1 - 1)) * (((P2^(Q2+1)) - 1) / (P2 - 1)) * (((P3^(Q3 + 1)) - 1) / (P3 - 1))

Below is implementation based on above formula.

`// C++ program to find sum of proper divisors for ` `// every element in an array. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `#define MAX 1000001 ` `#define pii pair<int, int> ` `#define F first ` `#define S second ` ` ` `// To store prime factors and their ` `// powers ` `vector<pii> factors[MAX]; ` ` ` `// Fills factors such that factors[i] is ` `// a vector of pairs containing prime factors ` `// (of i) and their powers. ` `// Also sets values in isPrime[] ` `void` `sieveOfEratothenese() ` `{ ` ` ` `// To check if a number is prime ` ` ` `bool` `isPrime[MAX]; ` ` ` `memset` `(isPrime, ` `true` `, ` `sizeof` `(isPrime)); ` ` ` `isPrime[0] = isPrime[1] = ` `false` `; ` ` ` ` ` `for` `(` `int` `i = 2; i < MAX; i++) ` ` ` `{ ` ` ` `// If i is prime, then update its ` ` ` `// powers in all multiples of it. ` ` ` `if` `(isPrime[i]) ` ` ` `{ ` ` ` `for` `(` `int` `j = i; j < MAX; j += i) ` ` ` `{ ` ` ` `int` `k, l; ` ` ` `isPrime[j] = ` `false` `; ` ` ` `for` `(k = j, l = 0; k % i == 0; l++, k /= i) ` ` ` `; ` ` ` `factors[j].push_back(make_pair(i, l)); ` ` ` `} ` ` ` `} ` ` ` `} ` `} ` ` ` `// Returns sum of proper divisors of num ` `// using factors[] ` `int` `sumOfProperDivisors(` `int` `num) ` `{ ` ` ` `// Applying above discussed formula for every ` ` ` `// array element ` ` ` `int` `mul = 1; ` ` ` `for` `(` `int` `i = 0; i < factors[num].size(); i++) ` ` ` `mul *= ((` `pow` `(factors[num][i].F, ` ` ` `factors[num][i].S + 1) - 1) / ` ` ` `(factors[num][i].F - 1)); ` ` ` `return` `mul - num; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `sieveOfEratothenese(); ` ` ` `int` `arr[] = { 8, 13, 24, 36, 59, 75, 91 }; ` ` ` `for` `(` `int` `i = 0; i < ` `sizeof` `(arr) / ` `sizeof` `(` `int` `); i++) ` ` ` `cout << sumOfProperDivisors(arr[i]) << ` `" "` `; ` ` ` `cout << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

7 1 36 55 1 49 21

This article is contributed by **Shubham Singh (singh_8)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Sum of all proper divisors of a natural number
- Find sum of divisors of all the divisors of a natural number
- Find all divisors of a natural number | Set 2
- Find all divisors of a natural number | Set 1
- Find the Nth digit in the proper fraction of two numbers
- Numbers with exactly 3 divisors
- Check if sum of divisors of two numbers are same
- Common Divisors of Two Numbers
- Sum of common divisors of two numbers A and B
- Count of numbers below N whose sum of prime divisors is K
- C++ Program for Common Divisors of Two Numbers
- Number of divisors of product of N numbers
- Find numbers with K odd divisors in a given range
- Sum of all prime divisors of all the numbers in range L-R
- Divide the two given numbers by their common divisors