//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;
}

#include <iostream>
using namespace std;

int main()
{
    cout << "Hello, World!" << 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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

int main()

#include "stdio.h"
#include <stdlib.h>

int main (int argCount, char** args) {
    int a = atoi(args[1]);
    int b = atoi(args[2]);

    unsigned int sum = 0;

    unsigned int p = 1;
    for (unsigned int i = 1; i < b; i++) {
        p = p * i;
    }

    // (b!, (1 + b)!, (2 + b)!, ..., (n + b)!)
    for (unsigned int i = 0; i < a; i++) {
        p = p * (i + b);
        sum = sum + p;
    }
    printf("y: %u\n", sum);
    return 0;
}