/*
Algorithm:
Step 1: Get radius of the cylinder from the user and store in variable r
Step 2: Get height of the cylinder from the user and store in variable h
Step 3: Multiply radius * radius * height * pi and store in v
Step 4: Display the volume
*/

#include <iostream>
using namespace std;


int main()
{
        float r; //define variable for radius
        float h; //define variable for height
        float v;
        float pi;
        pi=3.1416;

        cout<<"Enter radius:";
        cin>>r;

        cout<<"Enter height:";
        cin>>h;

        v=r*r*h*pi;  //compute volume

        cout<<"Radius:"<<r<<"\tHeight:"<<h<<endl; //display radius and height

        cout<<"\n************************\n";

        cout<<"Volume:"<<v<<endl;//display volume

        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 <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;
}

#include <iostream>
int main(){
    const char* const hello = "Hello, world!";
    
    const char* bruh = hello;
    
    char* const yeet = hello;
    
    std::cout << bruh << std::endl;
    
    std::cout << yeet << std::endl;
    
    return 0;
}

/*
    Place your bets!
    
    Will the program:
    a.) Print "Hello, world!" twice?
    b.) Compile error on line 5 (bruh initialize line) because the pointer gets implicit cast to non-const?
    c.) Compile error on line 7 (yeet initialize line) because the char gets implicit cast to non-const?
    d.) Both b and c?
    e.) Compile error line 11 (print yeet) because the pointer is constant and can't be incremented
    f.) Print "Hello, world!" then print the pointer address in hexadecimal
    g.) Both b and e?
    h.) Both c and e?
    i.) B, c, and e?
    
*/

// The answer is in this base 64 string:
// 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

// Iterative C++ program to
// implement Stein's Algorithm
//#include <bits/stdc++.h>
#include <bitset>
using namespace std;

// Function to implement
// Stein's Algorithm
int gcd(int a, int b)
{
    /* GCD(0, b) == b; GCD(a, 0) == a,
    GCD(0, 0) == 0 */
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    /*Finding K, where K is the
    greatest power of 2
    that divides both a and b. */
    int k;
    for (k = 0; ((a | b) & 1) == 0; ++k)
    {
        a >>= 1;
        b >>= 1;
    }

    /* Dividing a by 2 until a becomes odd */
    while ((a & 1) == 0)
        a >>= 1;

    /* From here on, 'a' is always odd. */
    do
    {
        /* If b is even, remove all factor of 2 in b */
        while ((b & 1) == 0)
            b >>= 1;

        /* Now a and b are both odd.
        Swap if necessary so a <= b,
        then set b = b - a (which is even).*/
        if (a > b)
            swap(a, b); // Swap u and v.

        b = (b - a);
    } while (b != 0);

    /* restore common factors of 2 */
    return a << k;
}

// Driver code
int main()
{
    int a = 12, b = 780;
    printf("Gcd of given numbers is %d\n", gcd(a, b));
    return 0;
}

#include <iostream>
using namespace std;

int main() {

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

}