Algorithms

nootanghimire wrote:

prediction for just half - one hour ahead of current time. In that case, is it possible to obtain near-to accurate prediction using atmospheric data of about 5 days prior ?

std::string mystdarr[][2] = {{"Abel","Hirst"},{"Benjamin","Foster"},{"Letty","Johnson"},{"George","Hanson"},{"Chris","Johnson"},{"Priscilla","Stein"}, {"",""}};
int arrsize = sizeof(mystdarr)/sizeof(mystdarr[0]);
for (auto it=0; it < arrsize - 1; ++it) {
   auto itFound = std::find_if(&mystdarr[it+1], &mystdarr[arrsize-1], [&](const std::string (&cmp)[2]) { return !cmp[1].compare(mystdarr[it][1]); });
   if (itFound != (std::string (*)[2])(mystdarr[arrsize-1]))
      std::cout << mystdarr[it][0].c_str() << " and " << (*itFound)[0].c_str() << " have the same family name (" << (*itFound)[1].c_str() << ")\n";
}

struct SS
{
   std::string firstName;
   std::string lastName;
   SS(const std::string& fname, const std::string& lname) : firstName(fname), lastName(lname) {};
};
std::vector<SS> myarr;
myarr.push_back(SS("Abel", "Hirst"));
myarr.push_back(SS("Benjamin", "Foster"));
myarr.push_back(SS("Letty", "Johnson"));
myarr.push_back(SS("George", "Hanson"));
myarr.push_back(SS("Chris", "Johnson"));
myarr.push_back(SS("Priscilla", "Stein"));
for (auto it=myarr.begin(); it!=myarr.end(); ++it)
{
   auto itFound = std::find_if(it + 1, myarr.end(), [&](const SS& cmp) {return !cmp.lastName.compare(it->lastName);});
   if (itFound != myarr.end())
      std::cout << it->firstName.c_str() << " and " << itFound->firstName.c_str() << " have the same family name (" << it->lastName.c_str() << ")\n";
}

Member 9963602 wrote:

I am a complete novice and have little to know idea what i am talking about so try not to be too wordy in your explinations.