problems with Recursion - C++ - c++

Question

Given a natural number n (1 <= n <= 500000), please output the summation of all its proper divisors.
Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.
e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
<<--This is a challenge i am trying to do and i am using recursion
int find_sum(std::vector <int> nums,long int sum,int num_now,long int j)
{
if(j<nums[num_now])
{
if(nums[num_now]%j==0)
{
sum=sum+j;
}
return find_sum(nums,sum,num_now,j+1);
}
else
{
return sum;
}
}
sum is the sum of all divisors,nums is the vector i stored number in,num_now is current member in vector,int j is 1 i use it to search for dividers,sadly using this i cant use numbers like 500000 it give's me error,is there any better way to do it or have i made a mistake somewhere.
--Thank you for your time

Answer 1

Here is a recursive way to solve your problem:
int find_sum(int x, int i)
{
if(i == 0)
return 0;
if(x % i == 0)
return i + find_sum(x, (i-1));
return find_sum(x, (i-1));
}
You need to call find_sum(N, N-1); in order to find sum of dividers of N (i must be less than given N because of strict inequality).
In your case it would be find_sum(20, 19);
e.g. my function returns:
71086 for N = 50000
22 for N = 20
0 for N = 1

Answer 2

I don't see the reason why you need to use recursion for solving this problem. I would prefer a more staightforward way to solve it.
long CalculateSumOfDivisors(int number)
{
long sum = 0;
for(int i=1; i<number; i++)
{
// If the remainder of num/i is zero
// then i divides num. So we add it to the
// current sum.
if(number%i==0)
{
sum+=i;
}
}
return sum;
}
Furthermore, we could write a more optimal algorithm, if we note the following:
Let that we have a number n and d is the smallest divisor of n that is greater of 1. (Apparently if the number n is a prime number there is any such a divisor). Then the larget divisor of n is the number n/d.
Based on this we can formulate a more optimal algorithm.
long CalculateSumOfDivisors(int number)
{
int smallestDivisor = FindSmallestDivisor(number);
if(smallestDivisor==1) return 1;
long sum = smallestDivisor;
// Calculate the possible greatest divisor.
int possibleGreatestDivisor = (int)floor(number/smallestDivisor);
for(int i=smallestDivisor+1; i<=possibleGreatestDivisor; i++)
{
if(number%i==0)
{
sum+=i;
}
}
return sum;
}
int FindSmallestDivisor(int number)
{
int smallestDivisor = 1;
for(int i=2; i<number; i++)
{
if(number%i==0)
{
smallestDivisor = i;
break;
}
}
return smallestDivisor;
}

Answer 3

I tried writing code with main function asking user to give the it wants to get sum of. here is the code , hope it helps.
#include<iostream>
using namespace std;
int Sum(int min, int max, int &val, int &sum){
if(min >= max)
return 0;
for ( ; min < max; min++){
if ( val%min == 0){
sum += min + val/min;
return Sum(++min,val/min, val,sum);
}
}
return 0;
}
int main(){
int s=1;
int val;
cout <<"Enter Val to sum:";
cin >> val;
Sum(2,val,val,s);
cout <<"Sum is :"<<s<<endl;
return 0;
}
Here Sum function is used recursively and passed parameters as shown in the code.
Hope it helps.

Answer 4

I don't think you should use recursion.
Instead start by looping from 1..N-1
When you find a divisor adjust the end value for the loop. Example if 2 is a divisor then you know N/2 is also a divisor. And just as important you know there can be no further divisors in the range ]N/2:N[. Likewise if 3 is a divisor then you know N/3 is also a divisor and you know there are no more divisors in the range ]N/3:N[.
Following that concept you can reduced the number of loops significantly for most numbers.
Something like:
long find_sum(int num)
{
long sum = 0;
int max = num;
int i = 1;
while(i < max)
{
if(num % i == 0)
{
sum += i; // Add i to the sum
max = num / i; // Decrement max for performance
if (max != i && max != num)
{
sum += max; // Add max when max isn't equal i
}
}
i++;
}
return sum;
}
Example:
num = 10
sum = 0
i = 1 -> sum = 1, max = 10
i = 2 -> sum = 1+2+5, max = 5
i = 3 -> sum = 1+2+5, max = 5
i = 4 -> sum = 1+2+5, max = 5
i = 5 -> return 8 (1+2+5)
num = 64
sum = 0
i = 1 -> sum = 1, max = 64
i = 2 -> sum = 1+2+32, max = 32
i = 3 -> sum = 1+2+32, max = 32
i = 4 -> sum = 1+2+32+4+16, max = 16
i = 5 -> sum = 1+2+32+4+16, max = 16
i = 6 -> sum = 1+2+32+4+16, max = 16
i = 7 -> sum = 1+2+32+4+16, max = 16
i = 8 -> sum = 1+2+32+4+16+8, max = 8
i = 9 -> return (1+2+32+4+16+8)
The number of loops are kept down by changing max whenever a new divisor is found.