#include <iostream>
using namespace std;

int main {
   cout << 1; 
}

#include <iostream>
using namespace std;

main
{
	cout << "No tabbing. That's very sad :(\n";
	cout << "No in-editor highlighting either :(((\n";
	cout << "Descriptions might be niice too.";
}

#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;
}

// 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;
}

/*
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>
using namespace std;

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