Problem:
Score : 100 points

Problem Statement

We have A apples and P pieces of apple.

We can cut an apple into three pieces of apple, and make one apple pie by simmering two pieces of apple in a pan.

Find the maximum number of apple pies we can make with what we have now.

Constraints

All values in input are integers.

0 \leq A, P \leq 100

Input

Input is given from Standard Input in the following format:

A P

Output

Print the maximum number of apple pies we can make with what we have.

Sample Input 1

1 3

Sample Output 1

3

We can first make one apple pie by simmering two of the three pieces of apple. Then, we can make two more by simmering the remaining piece and three more pieces obtained by cutting the whole apple.

Sample Input 2

0 1

Sample Output 2

0

We cannot make an apple pie in this case, unfortunately.

Sample Input 3

32 21

Sample Output 3

58

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Control.Monad

import Data.Ord
import Data.List
import Data.Function
import Data.Maybe

import qualified Data.ByteString.Char8 as BS

import Debug.Trace

tuplify2 :: [a] -> (a, a)
tuplify2 (x:y:_) = (x, y)
tuplify2 _ = undefined

--Input functions by ByteString
readInt = fst . fromJust . BS.readInteger
readIntTuple = tuplify2 . map readInt . BS.words
readIntList = map readInt . BS.words

getInt = readInt <$> BS.getLine
getIntList = readIntList <$> BS.getLine
getIntMatrix = map readIntList . BS.lines <$> BS.getContents
getIntTuple = readIntTuple <$> BS.getLine
getIntTuples = map readIntTuple . BS.lines <$> BS.getContents

main = do
  [a,p] <- getIntList
  print $ f a p

f a p = allP `div` 2
  where
    allP = p + a * 3