terminate called after throwing an instance of ‘std::out_of_range’

It looks to me as if this is due to a typo, and you should use the variable ‘j’ in the second loop. After the first loop,

for (i = 0; i < 52; i++)
{
    nums.push_back(i);
}

the variable ‘i’ contains the value 52, so it sounds expected that calling nums.at(i) would throw a std::out_of_range, since nums only contains 52 values, starting at index 0.

for(int j = 0; j < 52; j++)
{
    cout << nums.at(i) << "\n";
}

Fix it by replacing the argument of at() with ‘j’, which I assume was the original intent:

for(int j = 0; j < 52; j++)
{
    cout << nums.at(j) << "\n";
}