#include <iostream>
#include <fstream>
#include <string>
#include <cstring>

using namespace std;

//This program makes a new text file that contains all combinations of two letters.
// aa, ab, ..., zy, zz

int main(){
	string filename = "two_letters.txt";

	ofstream outFile;
	outFile.open(filename.c_str());

	if(!outFile.is_open()){
		cout << "Something's wrong. Closing..." << endl;
		return 0;
	}

	for(char first = 'a'; first <= 'z'; first++){
		for(char second = 'a'; second <= 'z'; second++){
			outFile << first << second << " ";
		}
		outFile << endl;
	}

	return 0;
}

#include <iostream>
#include <string>	//Should already be in iostream
#include <cstdlib>

//A word score adds up the character values. a-z gets mapped to 1-26 for the values of the characters.

//wordScore [wordValue]
//Pipe in the input into stdin, or type the words yourself.

//Lowercase words only
int characterValue(const char b){
	return ((b >= 'a') && (b <= 'z'))? ((b - 'a') + 1) : 0;
}

int main(int argc, char** argv){
    //The first argument specifies if you are trying to look for a certain word score
	int wordValue = (argc > 1)? std::atoi(argv[1]) : 0;

	std::string line;
	while(std::getline(std::cin, line)){
		int sum = 0;
		for(const char c : line){
			sum += characterValue(c);
		}
		if(wordValue){	//If wordValue is 0 or the sum is the correct value
			if(wordValue == sum){
				std::cout << line << std::endl;
			}
		} else {
			std::cout << sum << "\t" << line << std::endl;
		}
	}
	return 0;
}

//Constant prefix notation solver using bruh
//Could make it infix or postfix later
#include<string>
#include<vector>
#include<iostream>

std::vector<long double> bruhBuff;

long double operator ""bruh(long double a){
    bruhBuff.push_back(a);
    return a;
}

long double operator ""bruh(const char op){
    if(bruhBuff.size() < 2) throw "Bruh weak";
    
    long double b = bruhBuff.back();
    bruhBuff.pop_back();
    long double a = bruhBuff.back();
    bruhBuff.pop_back();
    
    
    switch(op){
        case (int)('+'):
            return a + b;
        case (int)('-'):
            return a - b;
        case (int)('*'):
            return a * b;
        case (int)('/'):
            return a / b;
    }
    return 69l;
}
 
int main(){
    1.0bruh;
    2.0bruh;
    std::cout << '+'bruh << std::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>
#include <cstring>
int main(int argc, char** argv){
	//With decimal
	if(strstr(argv[1], ".") != nullptr){
		int i = 0;
		//Skip i to first non 0 digit
		while(argv[1][i] < '1' || argv[1][i] > '9') ++i;
		//If digit comes before decimal
		if((argv[1] + i) < strstr(argv[1], ".")){	//Good example of pointer arithmetic
			std::cout << strlen(argv[1] + i) - 1 << std::endl;	//Another good example
		}else{
			//If digit is after decimal
			std::cout << strlen(argv[1] + i) << std::endl;
		}
	}else{
	//Without decimal
		int m = 0;
		int i = 0;
		while(argv[1][i] < '1' || argv[1][i] > '9') ++i;	//In case of some number like 0045
		for(; argv[1][i] != '\0'; ++i){
			if(argv[1][i] >= '1' && argv[1][i] <= '9') m = i + 1;
		}
		std::cout << m << std::endl;
	}
	return 0;
}

#include <iostream>
using namespace std;

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