standard deviation without mean c++ - c++

Question

I have a functioning program to find the standard deviation of many integers. However, I am to find a way to get the standard deviation without the mean.
I understand the formula is:
std dev = sqrt [(B - A^2/N)/N]
where
A is the sum of the data values;
B is the sum of the squared data values;
N is the number of data values.
but how would I write that in code?
This is my function for the deviation but it uses the mean:
float calculateSD(int arr[])
{
float sum = 0.0, mean, standardDeviation = 0.0;
int i;
for(i = 0; i < SIZE; ++i)
{
sum += arr[i];
}
mean = sum/SIZE;
for(i = 0; i < SIZE; ++i)
//convert standardDeviation to float
standardDeviation += static_cast<float>(pow(arr[i] - mean, 2));
//return standard deviation
return sqrt(standardDeviation / SIZE);
}

Answer 1

#include <iostream>
#include <vector>
#include <numeric>
#include <math.h>
double stddev(std::vector<int> const& data)
{
auto stats = std::make_pair(0.0,0.0);
stats = std::accumulate(data.begin(), data.end(), stats,
[](std::pair<double,double> stats, double x) {
stats.first += x;
stats.second += x * x;
return stats;
});
return sqrt((stats.second - pow(stats.first, 2.0) / data.size()) / data.size());
}
int main(int argc, const char *argv[])
{
std::cout << stddev({1,1,1,1}) << std::endl;
std::cout << stddev({1,2,1,2}) << std::endl;
std::cout << stddev({1,10,1,10}) << std::endl;
}