#include <iostream>
using namespace std;

int main()
{
    cout << "Hello, World!" << endl;
    return 0;
}

#include <iostream>
using namespace std;

/* Function: get_coeff
Parameters: double& coeff, int pos passed from bb_4ac
Return: type is void so no return, but does ask for user to input data that establishes what a b and c are.
*/
void get_coeff(double& coeff, int pos) {
  char position;
  if(pos == 1) {
    position = 'a';
  } else if(pos == 2) {   //a simple system to determine what coefficient the program is asking for.
    position = 'b';
  } else {
    position = 'c';
  }
  cout << "Enter the co-efficient " << position << ":"; //prompt to input coeff
  coeff = 5; //input coeff
}

/* Function: bb_4ac
Parameters: no parameters passed from main, but 3 params established in function, double a, b, c.
Return: b * b - 4 * a * c
*/
double bb_4ac() {
  double a, b, c; //coefficients of a quadratic equation
  get_coeff(a, 1); // call function 1st time
  get_coeff(b, 2); // call function 2nd time
  get_coeff(c, 3); // call function 3rd time
  return b * b - 4 * a * c; //return b * b - 4 * a * c
}

int main() {
  cout << "Function to calculate the discriminant of the equation. . . " << endl;
  double determinate = bb_4ac(); //assign double determinate to bb_4ac function
  cout << "The discriminant for given values is: " << determinate << endl;  //output the determinate!
}

//Leif Messinger
//Finds all sets of 5 5 letter words that don't have duplicate letters in either themselves or each other.
//First it reads the words in and puts them in groups of their bitmasks
//After that, we recurse on each group. Before doing that, we remove the group from the set of other groups to check it against.
#include <cstdio> //getchar, printf
#include <cassert> //assert
#include <vector>
#include <set>
#include <algorithm> //std::copy_if
#include <iterator> //std::back_inserter

#define CHECK_FOR_CRLF true
#define MIN_WORDS 5
#define MAX_WORDS 5
#define WORD_TOO_LONG(len) (len != 5)

const unsigned int charToBitmask(const char bruh){
	assert(bruh >= 'a' && bruh <= 'z');
	return (1 << (bruh - 'a'));
}

void printBitmask(unsigned int bitmask){
	char start = 'a';
	while(bitmask != 0){
		if(bitmask & 1){
			putchar(start);
		}

		bitmask >>= 1;
		++start;
	}
}

//Pointer needs to be deleted
const std::set<unsigned int>* getBitmasks(){
	std::set<unsigned int>* bitmasksPointer = new std::set<unsigned int>;
	std::set<unsigned int>& bitmasks = (*bitmasksPointer);

	unsigned int bitmask = 0;
	unsigned int wordLength = 0;
	bool duplicateLetters = false;
	for(char c = getchar(); c >= 0; c = getchar()){
		if(CHECK_FOR_CRLF && c == '\r'){
			continue;
		}
		if(c == '\n'){
			if(!(WORD_TOO_LONG(wordLength) || duplicateLetters)) bitmasks.insert(bitmask);
			bitmask = 0;
			wordLength = 0;
			duplicateLetters = false;
			continue;
		}
		if((bitmask & charToBitmask(c)) != 0) duplicateLetters = true;
		bitmask |= charToBitmask(c);
		++wordLength;
	}

	return bitmasksPointer;
}

void printBitmasks(const std::vector<unsigned int>& bitmasks){
	for(unsigned int bruh : bitmasks){
		printBitmask(bruh);
		putchar(','); putchar(' ');
	}
	puts("");
}

//Just to be clear, when I mean "word", I mean a group of words with the same letters.
void recurse(std::vector<unsigned int>& oldBitmasks, std::vector<unsigned int> history, const unsigned int currentBitmask){
	//If there's not enough words left
	if(oldBitmasks.size() + (-(history.size())) + (-MIN_WORDS) <= 0){
		//If there's enough words
		if(history.size() >= MIN_WORDS){
			//Print the list
			printBitmasks(history);
		}
		return;
	//To make it faster, we can stop it after 5 words too
	}else if(history.size() >= MAX_WORDS){
		//Print the list
		printBitmasks(history);
		return;
	}

	//Thin out the array with only stuff that matches the currentBitmask.
	std::vector<unsigned int> newBitmasks;
	std::copy_if(oldBitmasks.begin(), oldBitmasks.end(), std::back_inserter(newBitmasks), [&currentBitmask](unsigned int bruh){
		return (bruh & currentBitmask) == 0;
	});

	while(newBitmasks.size() > 0){
		//I know this modifies 'oldBitmasks' too. It's intentional.
		//This makes it so that the word is never involved in any of the child serches or any of the later searches in this while loop.
		const unsigned int word = newBitmasks.back(); newBitmasks.pop_back();

		std::vector<unsigned int> newHistory = history;
		newHistory.push_back(word);

		recurse(newBitmasks, newHistory, currentBitmask | word);
	}
}

int main(){
	const std::set<unsigned int>* bitmasksSet = getBitmasks();
	std::vector<unsigned int> bitmasks(bitmasksSet->begin(), bitmasksSet->end());
	delete bitmasksSet;

	recurse(bitmasks, std::vector<unsigned int>(), 0);

	return 0;
}

#include <iostream>
#include <cmath>
#include <string.h>

using namespace std;

int main() {

    string tickerName;
    int numOfContracts;
    float currentOptionValue;

    cout << "Enter a stock ticker: ";
    getline(cin, tickerName);

    cout << "Enter the current number of " << tickerName << " contracts you are holding: ";
    cin >> numOfContracts;


    cout << "Enter the current price of the option: ";
    cin >> currentOptionValue;


    cout << "The value of your " << tickerName << " options are: $" << (currentOptionValue * 100.00) * (numOfContracts);

    cout << endl;

    return 0;
}

/*
Good morning! Here's your coding interview problem for today.

This problem was asked by LinkedIn.

A wall consists of several rows of bricks of various integer lengths and uniform height. Your goal is to find a vertical line going from the top to the bottom of the wall that cuts through the fewest number of bricks. If the line goes through the edge between two bricks, this does not count as a cut.

For example, suppose the input is as follows, where values in each row represent the lengths of bricks in that row:

[[3, 5, 1, 1],
	[2, 3, 3, 2],
	[5, 5],
	[4, 4, 2],
	[1, 3, 3, 3],
	[1, 1, 6, 1, 1]]

The best we can we do here is to draw a line after the eighth brick, which will only require cutting through the bricks in the third and fifth row.

Given an input consisting of brick lengths for each row such as the one above, return the fewest number of bricks that must be cut to create a vertical line.

AUTHORS NOTE:
Makes following assumptions:
- Each row is same length
- Data is in file called "data.dat" and formatted in space-separated rows
- The cuts at the beginning and end of the wall are not solutions

This requires the following file named data.dat that is a space separated file, or similar formatted file:
----START FILE----
3 5 1 1
2 3 3 2
5 5
4 4 2
1 3 3 3
1 1 6 1 1
----END FILE----
*/

#include <algorithm>
#include <iostream>
#include <fstream>
#include <map>
#include <sstream>
#include <string>
#include <vector>
using namespace std;

int main()
{
	vector<vector<int>> wall;

	ifstream in;
	in.open("data.dat");
	if(!in.good())
	{
	cout << "ERROR: File failed to open properly.\n";
	}

	/* Get input from space separated file */
	string line;
	while(!in.eof())
	{
	getline(in, line);

	int i;
	vector<int> currv;
	stringstream strs(line);
	while(strs >> i)
	currv.push_back(i);
	wall.push_back(currv);
	}

	

	/* Convert each value from "length of brick" to "position at end of brick" */
	for(int y = 0; y < wall.size(); y++)
	{
	wall.at(y).pop_back(); //Delet last val
	for(int x = 1; x < wall.at(y).size(); x++) //Skip the first bc data doesn't need change
	wall.at(y).at(x) += wall.at(y).at(x-1);
	}

	/* Check output. COMMENT OUT */
	// for(auto row : wall)
	// {
	// for(int pos : row)
	// cout << pos << " ";
	// cout << endl;
	// }

	/* Determine which ending position is most common, and cut there */
	//Exclude final position, which will be the size of the wall

	int mode = -1;
	int amt = -1;
	vector<int> tried;
	for(auto row : wall)
	{
	for(int pos : row) //For each pos in the wall
	{
	//Guard. If pos is contained in the list, skip pos
	if(find(tried.begin(), tried.end(), pos) != tried.end())
	continue;
	tried.push_back(pos);

	/* Cycle through each row to see if it contains the pos */
	int curramt = 0;
	for(auto currrow : wall)
	{
	if( find( currrow.begin(), currrow.end(), pos ) != currrow.end() )
	curramt++;
	}
	//cout << pos << " " << curramt << endl; 

	if(curramt > amt)
	{
	amt = curramt;
	mode = pos;
	}
	}
	}

	cout << "Please cut at position " << mode << endl;
	cout << "This will cut through " << (wall.size() - amt) << " bricks." << endl;

	return 0;
}

#include <iostream>
using namespace std;

int main() {

cout << "Hello World!\n";
// Prints out "Hello World"
return 0;

}