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Introduction

It was just a normal day, here at your military base. Suddenly, you looked up to see a bomb falling down from the sky! You need to create a program to predict where it will fall, and tell you the location.

Challenge

You will be given a bomb in the form of Start-X Start-Y X-velocity Y-velocity. You can assume that they will all be positive integers. 0 is a valid input only for Start-x and Start-y. The bomb starts with x Start-x, and y Start-y. Each "frame" (no graphical output required), the bomb will move x-velocity units right, then y-velocity units down. Continue doing this until the bomb reaches y of 0 or below. Then, output what it's x was when it hit the ground.

Test Cases

1

Input:0 10 2 2

Output:10

2

Input:5 4 3 2

Output:11

3

Input:11,8,7,3

Output:32

This is , so shortest answer in bytes wins!

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  • 6
    \$\begingroup\$ Once you do out the math, it looks like the formula leaves little to golf. \$\endgroup\$ – xnor Apr 4 '17 at 21:23
  • \$\begingroup\$ @xnor Maybe the ceiling function could be golfed. But I can't see anywhere else. \$\endgroup\$ – fəˈnɛtɪk Apr 4 '17 at 21:24
  • \$\begingroup\$ You can assume that they will all be positive integers 0 is not a positive integer. Can you clarify if 0 is a valid input or not? \$\endgroup\$ – James Apr 4 '17 at 21:26
  • \$\begingroup\$ Looks like 0 is valid input from test case #1 \$\endgroup\$ – Timtech Apr 4 '17 at 21:27
  • \$\begingroup\$ 0 is a valid input only for start-x and start-y. editing now. \$\endgroup\$ – Baked Potato Apr 4 '17 at 21:28
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05AB1E, 9 bytes

¹³²(I÷(*+

Try online

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TI-Basic, 21 bytes

Input :Prompt A,B:X+A+Aint(Y/B-E~9

If the ceiling function did not have to be implemented, it would be more straightforward at 13 bytes:

Input :Prompt A,B:X+AY/B
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  • \$\begingroup\$ Don't you need to prompt for x and y as well? \$\endgroup\$ – fəˈnɛtɪk Apr 4 '17 at 21:28
  • \$\begingroup\$ @fəˈnɛtɪk In TI-83/84 BASIC, providing no arguments to Input is equivalent to Prompt X,Y, taking location from the graph screen. It's usually more of a savings when only one or two values are needed, although it saves 1 byte in this situation as well. \$\endgroup\$ – Timtech Apr 4 '17 at 22:03
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Python, 45 28 Bytes

lambda a,b,x,y:a+x*-(-b//y)

Try it online!

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  • \$\begingroup\$ The import has to be scored. meta post \$\endgroup\$ – mbomb007 Apr 4 '17 at 21:28
  • \$\begingroup\$ @mbomb007 Still, with the import is 45 bytes \$\endgroup\$ – Anthony Pham Apr 4 '17 at 21:28
  • \$\begingroup\$ Why not use your own ceil function? lambda a,b,x,y:a+x*int(b/y+.5) \$\endgroup\$ – James Apr 4 '17 at 21:32
  • \$\begingroup\$ @DJMcMayhem Just made one 2 bytes shorter than yours \$\endgroup\$ – Anthony Pham Apr 4 '17 at 21:48
  • \$\begingroup\$ a--b//y*x seems to work. \$\endgroup\$ – Ørjan Johansen Apr 5 '17 at 1:25
1
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Scheme 40 36 Bytes

λ(x y u v)(+ x(* u(ceiling(/ y v))))
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  • \$\begingroup\$ Does scheme require all of those spaces between identifiers and parentheses \$\endgroup\$ – Generic Display Name Apr 4 '17 at 21:57
  • \$\begingroup\$ Actually I can get rid of some of them...thanks \$\endgroup\$ – Penguino Apr 4 '17 at 22:00
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Haskell, 21 bytes

(a#b)x y=a-x*div(-b)y

The trivial answer: defines an operator that takes 4 arguments and then uses the formula.

Could probably be golfed more.

EDIT: Yep, Ørjan Johansen points out we can use a-x*div b(-y), which brings us to 22 bytes. Try it online!

At this point, the code is really more other people than mine, so I'm going to mark this answer community wiki. Feel free to golf this further if you want :)

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  • \$\begingroup\$ Ouch, importing Data.Ratio ... try a-x*div(-b)y. \$\endgroup\$ – Ørjan Johansen Apr 5 '17 at 1:28
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JavaScript, 29 28 bytes

Thanks @user5090812 for golfing off 1 byte

(a,b,x,y)=>a+x*((b+y-1)/y|0)

Try it online!

If it did not have to move x then y but was instead falling in a straight line it would be 18 bytes

(a,b,x,y)=>a+x*b/y
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    \$\begingroup\$ You could do (a,b,x,y)=>a+x*(b/y|0+1) and save 5 bytes \$\endgroup\$ – powelles Apr 4 '17 at 22:28
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    \$\begingroup\$ @powelles If I do that I will end up overshooting for exact values. \$\endgroup\$ – fəˈnɛtɪk Apr 4 '17 at 23:54
  • \$\begingroup\$ Weird. I expected x|0+1 to perform the same as Math.ceil and didn't actually test it. \$\endgroup\$ – powelles Apr 5 '17 at 0:30
  • \$\begingroup\$ @powelles if you have 1|0+1 you end up with 2. With Math.ceil you will get 1 \$\endgroup\$ – fəˈnɛtɪk Apr 5 '17 at 0:36
  • \$\begingroup\$ I mistakenly made the assumption that ceiling could be implemented as floor+1. \$\endgroup\$ – powelles Apr 5 '17 at 0:53
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C#, 46 bytes

(a,b,c,d)=>a+c*(int)Math.Ceiling((double)b/d);
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0
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Ruby, 21 bytes

->x,y,v,w{x-v*(y/-w)}
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0
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Java 7, 66 bytes

int f(int a,int b,int x,int y){return a+x*(int)Math.ceil(1f*b/y);}
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