Implementation of Sum of divisor algorithm in c++ - c++

Question

I was trying to implement the sum of divisors algorithm in c++ i.e.,
Let n = p1^a1 * p2^a2 * .... * pk^ak
Then
σ(n) = ( (p1^(a1+1) -
1) / (p1-1) ) * ( (p2^(a2+1) - 1) / (p2-1) ) * ... * ( (pk^(ak+1) - 1) / (pk-1) )
The function for prime factorization that pushes factor in a vector<int> p
void primeFactors(int n)
{
while (n%2 == 0)
{
p.push_back(2);
n = n/2;
}
for (ull i = 3; i <= sqrt(n); i = i+2)
{
while (n%i == 0)
{
p.push_back(i);
n = n/i;
}
}
if (n > 2)
p.push_back(n);
}
Then I create a copy of vector<int> p in vector<int> pk as vector<int> pk(p) and do following to have only unique elements in p.
auto it = unique(p.begin(),p.end());
p.resize(distance(p.begin(),it) );
Now the final to get sum as per the formula:
for(int i=0;i<p.size();i++){
sum *= (pow(p[i],count(pk.begin(), pk.end(), p[i]))-1)/(p[i]-1);
}
The above implemention is wrong as I am not getting correct output. Where am I making a mistake? Is their a better method for implementing it?

Answer 1

Your final loop for calculating the result is pi^ai instead of pi^(ai+1). Change it to this:
for (int i = 0; i < p.size(); i++){
sum *= (pow(p[i], count(pk.begin(), pk.end(), p[i]) + 1) - 1) / (p[i] - 1);
}
One more thing, instead of using vectors and using unique and count you can simply use a map<int, int> and instead of getting the number of occurrences of each prime number in O(n) get it in O(lgn) like this:
void primeFactors(int n, map<int, int> &facs){
facs.clear();
for (long long p = 2; p*p <= n; p++){
while (n%p == 0){
facs[p] ++;
n /= p;
}
}
if (n > 1)
facs[n] = 1;
}
long long getRes(map<int, int> &facs){
long long t, res = 1;
for (auto it = facs.begin(); it != facs.end(); ++it){
t = pow(it->first, it->second + 1) - 1;
t /= (it->first - 1);
res *= t;
}
return res;
}
In the above code I also used pow function for calculating the result. We can eliminate this function by storing power values instead of count in the map, like this:
void primeFactors(int n, map<int, long long> &facs){
facs.clear();
for (long long p = 2; p*p <= n; p++){
while (n%p == 0){
if (facs.count(p))//this way we can save pi^(ai+1) instead of counting prime numbers
facs[p] *= p;
else
facs[p] = p*p;
n /= p;
}
}
if (n > 1)
facs[n] = n*n;
}
long long getRes(map<int, long long> &facs){
long long t, res = 1;
for (auto it = facs.begin(); it != facs.end(); ++it){
t = it->second - 1; //this is for pi^(ai+1)-1
t /= (it->first - 1);
res *= t;
}
return res;
}