//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;
}

// 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;
/*
 Description: uses switch case statements to determine whether it is hot or not outside.
 Also uses toupper() function which forces user input char to be uppercase in order to work for the switch statement
*/

int main() {
  char choice;
  cout << "S = Summer, F = Fall, W = Winter, P = Spring" << endl;
  cout << "Enter a character to represent a season: ";asdasdasdasd
  cin >> choice;

  enum Season {SUMMER='S', FALL='F', WINTER='W', SPRING='P'};

  switch(toupper(choice))  // This switch statement compares a character entered with values stored inside of an enum
  {
  case SUMMER:
    cout << "It's very hot outside." << endl;
    break;
  case FALL:
    cout << "It's great weather outside." << endl;
    break;
  case WINTER:
    cout << "It's fairly cold outside." << endl;
    break;
  case SPRING:
    cout << "It's rather warm outside." << endl;
    break;
  default:
    cout << "Wrong choice" << endl;
    break;
  }
  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 Stripe.

Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well.

For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.

You can modify the input array in-place.
*/
#include <iostream>
using namespace std;

int calcMissing(int* input, int size)
{
	int sum = 0;
	int n = 1; //add one to account for missing value
	for(int i = 0; i < size; i++)
	{
	if(input[i] > 0)
	{
	sum += input[i];
	n++;
	}
	}

	//If no numbers higher than 0, answer is 1
	if(sum == 0)
	return 1;

	return (n*(n+1)/2) - sum; //Formula is expectedSum - actualSum
	/* expectedSum = n*(n+1)/2, the formula for sum(1, n) */
}

int main()
{

	cout << calcMissing(new int[4]{3, 4, -1, 1}, 4) << endl;
	cout << calcMissing(new int[3]{1, 2, 0}, 3) << endl;

	//No positive numbers
	cout << calcMissing(new int[1]{0}, 1) << endl;
}

#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;
}