Just another day in the world of Minecraft, Steve is getting ready to start his next exciting project -- building a railway system!
![Alt text](http://static.planetminecraft.com/files/resource_media/screenshot/1506/2015-02-02_0156508603066.jpg)
But first, Steve needs to melt some iron ores in the furnace to get the main building blocks of rails -- iron ingots. ![Alt text](http://www.blocksandgold.com/media/catalog/product/cache/3/image/35x/cbcbef48e5e3bcce7c7ed908f20bc5b4/i/r/ironingot_icon32.png)
Each iron ingot takes 11 seconds* to produce. Steve needs a lot of them, and he has the following fuel options to add into the furnace: 
* Buckets of lava, each lasts 800 seconds* ![Alt text](http://www.minecraftguides.org/images/items/bucketL.gif)
* Blaze rod, each lasts 120 seconds ![Alt text](http://www.minecraftguides.org/images/items/blazerod.gif)
* Coals, each lasts 80 seconds ![Alt text](http://www.minecraftguides.org/images/items/coal.gif)
* Blocks of Wood, each lasts 15 seconds ![Alt text](http://www.minecraftguides.org/images/itemGifs/logs.gif)
* Sticks, each lasts 1 second* ![Alt text](http://www.minecraftguides.org/images/items/stick.gif)
In Ruby: 
Write a function `calc_fuel` that calculates the **minimum** amount of fuel needed to produce a certain number of iron ingots. This function should return a hash of the form `{:lava => 2, :blaze_rod => 1, :coal => 1, :wood => 0, :stick => 0}`
In JavaScript:
Write a function `calcFuel` that calculates the **minimum** amount of fuel needed to produce a certain number of iron ingots. This function should return an object of the form `{lava: 2, blazeRod: 1, coal: 1, wood: 0, stick: 0}`
In Python:
Write a function `calc_fuel` that calculates the **minimum** amount of fuel needed to produce a certain number of iron ingots. This function should return a dictionary of the form `{"lava": 2, "blaze rod": 1, "coal": 1, "wood": 0, "stick": 0}`
---
*fictional values
To all the Minecraft players out there: 
feel free to expand this series or let me know if you have any ideas related to Minecraft that can be turned into codewars puzzles. Some ideas I have that might potentially be turned into katas:
* distance traveled in real world vs. in Nether
* shortest path problems related to mining diamonds/gold/goodies that appears in different levels under ground
* growth of animal population from breeding
* redstone stuff?!
If you do end up expanding this series, please send me a link of your kata so I can check it out and include a link to your kata here :-)
* [Minecraft Series #1: Steve wants to build a beacon pyramid](https://www.codewars.com/kata/minecraft-series-number-1-steve-wants-to-build-a-beacon-pyramid/ruby)
* [Minecraft Series #3: Lava is amazing! ](https://www.codewars.com/kata/583a23d40cf946ec380002c2)
* [Minecraft Series #4: Lava is amazing, however...](https://www.codewars.com/kata/583a6b0b171f3a3c3f0003e3)
def calc_fuel(n):
    dic={'lava':0,'blaze rod':0,'coal':0,'wood':0,'stick':0}
    time=n*11
    poss=[800,120,80,15,1]
    i=0
    while time>0:
        while time-poss[i]>=0:
            time-=poss[i]
            if poss[i]==800:
                dic['lava']+=1
            elif poss[i]==120:
                dic['blaze rod']+=1
            elif poss[i]==80:
                dic['coal']+=1
            elif poss[i]==15:
                dic['wood']+=1
            elif poss[i]==1:
                dic['stick']+=1
        
        i+=1
        
    return dic
        
            
            
        
    
        
            
            
            
            
            
            
            
            
            
            

def calc_fuel(n):
    dic={'lava':0,'blaze rod':0,'coal':0,'wood':0,'stick':0}
    time=n*11
    poss=[800,120,80,15,1]
    i=0
    while time>0:
        while time-poss[i]>=0:
            time-=poss[i]
            if poss[i]==800:
                dic['lava']+=1
            elif poss[i]==120:
                dic['blaze rod']+=1
            elif poss[i]==80:
                dic['coal']+=1
            elif poss[i]==15:
                dic['wood']+=1
            elif poss[i]==1:
                dic['stick']+=1
        
        i+=1
        
    return dic
        
            
            
        
    
        
            
            
            
            
            
            
            
            
            
            
