AOA friends please help me to solve the program of C++

#include <iostream>
using namespace std;

int main() {
int passengers = 126;
int empty_seats = 0;
//passengers =126;
passengers after 1st bas leaving= passengers - 50;
passengers after 2nd bas leaving= passengers - 50;
passengers after 3rd bas= passengers;
empty_seats= 50 % passengers;
cout << "Empty seats in last bas"<< empty_seats <<endl;

return 0;
}

Look up A Star Path Search. Maybe you can even solve this with the Bus Against The Wall algorithm.

This is hardly a programming problem. It’s a math problem. Do the math first, then write the code.

Your current solution a) has too many steps (it requires exactly one step), and b) is implemented backwards.

(Also, you’ve misspelled “bus” four out of four times!)

@nitwhiz That's mean :D

One liner

int empty_seats = input % 50 == 0 ? 0 : input <= 50 ?
50 - input : 50 - (input % 50);

If you're too stupid to solve this, just quit programming as career option and try flipping burgers, maybe that's something you can do.

@iiii "n%50 && (50 - n%50)" is either 0 as intended if n is divisble by zero, but otherwise, the result is 1 like with any boolean expression.

And "(50-n)%50" won't work as intended for n>50.

You can get the number of passengers on the last bus by calculating the number of passengers total modulo 50 since modulo gives you the number “left over” when one number is divided by another and doesn’t divide evenly. (Eg: 5 mod 2 is 1 since 2/2 = 4, leaving 1 left over)

Then to calculate the number of empty seats on the last bus, you subtract that value from 50.

It’s easy to overthink these types of problems, especially if you don’t have a solid grasp of mathematics to begin with (which I don’t necessarily think is a requirement for programming - though it definitely makes things easier).

That being said, no idea what AOA means, so sorry if you were asking a group of people that doesn’t include me. 😅

Typing on a phone is annoying.

int lastBus(int passengers) {
int emptySeats = 50 - (passengers % 50);
if (emptySeats == 50) {
return 0;
}
return emptySeats;
}

@iiii It does work as specified in the standard: "(a/b) * b + a%b shall equal a".

For e.g. (-60)%50, that's a=-60 and b=50. a/b is -1, times 50 is -50. The only thing that you can add to -50 and get -60 as result is -10, so:

(-60)%50 == -10.

That's because % isn't actually modulo, it's the remainder.

@iiii Then again, JavaScript isn't quite C++. ^^

@iiii Not particularly helpful if the ask is for help writing a program in C++, friend.

Doesn’t modulo work the same way in C++ and JavaScript? They both return the remainder...

@iiii That's because the % operator isn't a modulo operator in C/C++. It's the remainder, not modulo.

@iiii Eh, I think the whole “actually returning a boolean from a logical comparison” rather than “returning one of the expressions from a logical comparison” makes more sense, honestly.

@iiii C/C++ do it their way because it gives the compiler more leeway for optimisation. E.g. the actual calculation doesn't have to be computed if the compiler can do something equivalent under the "as-if-rule".

r/homeworkwhores 🤷‍♂️

@iiii Just because you _can_ do things in one line doesn’t mean you _should_, friend.

@Fast-Nop @iiii Google "eevee stop reinventing C" and go to the section about modulo if you want a good rundown of that whole fiasco