# Euler problems/191 to 200

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− | A brief tutorial on solving this problem is available [http://whttp://maztravel.com/haskell/euler_problem_191.html here] |
+ | A brief tutorial on solving this problem is available [http://maztravel.com/haskell/euler_problem_191.html here] |

## Revision as of 17:58, 30 April 2008

## Problem 191

Prize Strings

A couple of notes. I was too lazy to memoize this, so I just ran it twice, once with 15 and then again with 30. I pasted the output of the 15 run into the code. The way to get a handle on this is to just case it out. Ask yourself what can I add to award (n-1) to generate award (n). You can add an O to the end of all of award (n-1). You can add an L to any award (n-1) that doesn't contain an L, and you can add an A to award (n-1) provided it doesn't end with two A's. So the function hasM_LsAndEndsInN_As is just what is needed to cover all of the cases. Henry Laxen April 29, 2008

award 1 = 3 award 15 = 107236 award k = award (k-1) -- + O + sum [ hasM_LsAndEndsInN_As 0 i (k-1) | i<-[0..2] ] -- +L + sum [ hasM_LsAndEndsInN_As i j (k-1) | i<-[0,1], j<-[0,1] ] -- +A hasM_LsAndEndsInN_As 0 0 1 = 1 -- O hasM_LsAndEndsInN_As 1 0 1 = 1 -- L hasM_LsAndEndsInN_As 0 1 1 = 1 -- A hasM_LsAndEndsInN_As _ _ 1 = 0 hasM_LsAndEndsInN_As 0 0 15 = 5768 hasM_LsAndEndsInN_As 0 1 15 = 3136 hasM_LsAndEndsInN_As 0 2 15 = 1705 hasM_LsAndEndsInN_As 1 0 15 = 54736 hasM_LsAndEndsInN_As 1 1 15 = 27820 hasM_LsAndEndsInN_As 1 2 15 = 14071 hasM_LsAndEndsInN_As m n k | m < 0 || n < 0 = 0 | n == 0 = sum [ hasM_LsAndEndsInN_As (m-1) i (k-1) | i<-[0..2]] -- +L + sum [ hasM_LsAndEndsInN_As m i (k-1) | i<-[0..2]] -- +O | n 0 = hasM_LsAndEndsInN_As m (n-1) (k-1) -- + A -- Count awards of length k that have "m" L's in them, and end in "n" A's problem191 n = do let p a b c d = "hasM_LsAndEndsInN_As " ++ foldl (\x y -> x ++ (show y) ++ " ") "" [a,b,c] ++ "= " ++ (show d) putStrLn $ "award " ++ (show n) ++ " = " ++ show (award n) mapM_ (\(i,j) -> putStrLn $ p i j n (hasM_LsAndEndsInN_As i j n)) [ (i,j) | i<-[0..1], j<-[0..2]]

A brief tutorial on solving this problem is available here