# Daily Challenge #69 - Going to the Cinema

Today's challenge is to help John make an important financial decision!

John likes to go to the cinema, but he wants to find the most cost-effective way to go. He can buy a ticket for $15, or he can buy a membership card for$500. Every time he uses the membership card, the ticket will be 0.9 times the price he paid for the last one.

Ex. If John goes to the cinema three times:
A : 15 * 3 = 45
B : 500 + 15 * 0.90 + (15 * 0.90) * 0.90 + (15 * 0.90 * 0.90) * 0.90 ( = 536.5849999999999, no rounding for each ticket)

Create a function movie with three parameters: card (price of the card), ticket(normal price of a ticket), perc(fraction of what he paid for the previous ticket) and returns the first n such that ceil(price of B) < price of A

More examples:
movie(500, 15, 0.9)should return43
(with card the total price is 634, with tickets 645)

movie(100, 10, 0.95) should return 24
(with card the total price is 235, with tickets 240)

Good luck!

Today's challenge comes from CodeWars user g964. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

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### Discussion

Here is my try:

function movie(card, ticket,perc){
let i = 0
let ticketFullPrice = 0
let lowedTicket = ticket

while(card>ticketFullPrice){
ticketFullPrice += ticket
lowedTicket *= perc
card += lowedTicket
i++
}
return i
}
movie(500,15,0.9)
movie(100, 10, 0.95)


"first" lmao

const movie = (card, ticket, part) => {
let i = 0
let a = 0
let b = card
do {
a += ticket
b += ticket * part ** i
i++
} while (Math.ceil(b) >= a)
return i - 1
}


an "optimization" (no power)

const movie = (card, ticket, part) => {
let i = 0
let a = 0
let b = card
let cheap = ticket
while (true) {
a += ticket
b += cheap
cheap *= part
if (Math.ceil(b) < a) return i
i++
}
}


"optimization II" (no Math.ceil, comparison with 0 instead of 1)

const movie = (card, ticket, part) => {
let i = 0
let dif = -card - 1
let cheap = ticket
while (true) {
dif += ticket - cheap
cheap *= part
if (dif > 0) return i
i++
}
}


idiocy.

const movie = (card, ticket, part) => {
let i = 1
let dif = -card - 1
let cheap = ticket
while ((dif += ticket - (cheap *= part)) <= 0) i++
return i
}


denied on first lol

Check timestamp by hovering the date lol

Elm

module Movie exposing (movie)

getReduction : Int -> Float -> Float
getReduction times percentage =
percentage ^ toFloat times

profitableNumberOfMovies : Int -> Int -> Float -> Int -> Int
profitableNumberOfMovies card ticket percentage times =
let
reduction : Float
reduction =
getReduction times percentage

increasedCard : Int
increasedCard =
ceiling <| toFloat card + toFloat ticket * reduction

increasedTicket : Int
increasedTicket =
ticket * times
in
if increasedCard < increasedTicket then
times - 1

else
profitableNumberOfMovies increasedCard ticket percentage <| times + 1

movie : Int -> Int -> Float -> Int
movie card ticket percentage =
profitableNumberOfMovies card ticket percentage 1


Tests

module MovieTest exposing (suite)

import Expect
import Movie exposing (movie)
import Test exposing (Test)

suite : Test
suite =
Test.describe "Movie"
[ Test.test "It should return 43 for movie 500 15 0.9" <|
\_ ->
Expect.equal 43 <| movie 500 15 0.9
, Test.test "It should return 24 for movie 100 10 0.95" <|
\_ ->
Expect.equal 24 <| movie 100 10 0.95
]


#!/usr/bin/perl
use warnings;
use strict;

sub movie {
my ($card,$ticket, $perc) = @_; my$reduced = my $sum =$ticket * $perc; my$count = 0;
$sum += ($reduced *= $perc) while$ticket * ++$count <$card + int $sum; return$count
}

use Test::More tests => 3;

is movie(100, 200, 0.5),   1;
is movie(500,  15, 0.9),  43;
is movie(100,  10, 0.95), 24;


JavaScript

const movie = (card, ticket, perc) => {
let n = 1;
let frac = ticket * perc;
let a = ticket * n;
let b = card + frac;
while(a < Math.ceil(b)) {
n++;
frac = frac * perc;
b = b + frac;
a = ticket * n;
}
return n;
}