Difference between revisions of "Euler problems/111 to 120"
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| + | Do them on your own! |
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| − | == [http://projecteuler.net/index.php?section=problems&id=111 Problem 111] == |
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| − | Search for 10-digit primes containing the maximum number of repeated digits. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | import Control.Monad (replicateM) |
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| − | |||
| − | -- All ways of interspersing n copies of x into a list |
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| − | intr :: Int -> a -> [a] -> [[a]] |
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| − | intr 0 _ y = [y] |
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| − | intr n x (y:ys) = concat |
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| − | [map ((replicate i x ++) . (y :)) $ intr (n-i) x ys |
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| − | | i <- [0..n]] |
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| − | intr n x _ = [replicate n x] |
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| − | |||
| − | -- All 10-digit primes containing the maximal number of the digit d |
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| − | maxDigits :: Char -> [Integer] |
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| − | maxDigits d = head $ dropWhile null |
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| − | [filter isPrime $ map read $ filter ((/='0') . head) $ |
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| − | concatMap (intr (10-n) d) $ |
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| − | replicateM n $ delete d "0123456789" |
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| − | | n <- [1..9]] |
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| − | |||
| − | problem_111 = sum $ concatMap maxDigits "0123456789" |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=112 Problem 112] == |
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| − | Investigating the density of "bouncy" numbers. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | isIncreasing' n p |
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| − | | n == 0 = True |
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| − | | p >= p1 = isIncreasing' (n `div` 10) p1 |
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| − | | otherwise = False |
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| − | where |
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| − | p1 = n `mod` 10 |
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| − | |||
| − | isIncreasing :: Int -> Bool |
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| − | isIncreasing n = isIncreasing' (n `div` 10) (n `mod` 10) |
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| − | |||
| − | isDecreasing' n p |
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| − | | n == 0 = True |
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| − | | p <= p1 = isDecreasing' (n `div` 10) p1 |
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| − | | otherwise = False |
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| − | where |
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| − | p1 = n `mod` 10 |
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| − | |||
| − | isDecreasing :: Int -> Bool |
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| − | isDecreasing n = isDecreasing' (n `div` 10) (n `mod` 10) |
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| − | |||
| − | isBouncy n = not (isIncreasing n) && not (isDecreasing n) |
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| − | nnn=1500000 |
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| − | num150 =length [x|x<-[1..nnn],isBouncy x] |
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| − | p112 n nb |
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| − | | fromIntegral nnb / fromIntegral n >= 0.99 = n |
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| − | | otherwise = prob112' (n+1) nnb |
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| − | where |
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| − | nnb = if isBouncy n then nb + 1 else nb |
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| − | |||
| − | problem_112=p112 (nnn+1) num150 |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=113 Problem 113] == |
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| − | How many numbers below a googol (10100) are not "bouncy"? |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | import Array |
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| − | |||
| − | mkArray b f = listArray b $ map f (range b) |
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| − | |||
| − | digits = 100 |
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| − | |||
| − | inc = mkArray ((1, 0), (digits, 9)) ninc |
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| − | dec = mkArray ((1, 0), (digits, 9)) ndec |
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| − | |||
| − | ninc (1, _) = 1 |
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| − | ninc (l, d) = sum [inc ! (l-1, i) | i <- [d..9]] |
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| − | |||
| − | ndec (1, _) = 1 |
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| − | ndec (l, d) = sum [dec ! (l-1, i) | i <- [0..d]] |
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| − | |||
| − | problem_113 = sum [inc ! i | i <- range ((digits, 0), (digits, 9))] |
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| − | + sum [dec ! i | i <- range ((1, 1), (digits, 9))] |
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| − | - digits*9 -- numbers like 11111 are counted in both inc and dec |
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| − | - 1 -- 0 is included in the increasing numbers |
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| − | </haskell> |
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| − | Note: inc and dec contain the same data, but it seems clearer to duplicate them. |
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| − | |||
| − | it is another way to solution this problem: |
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| − | <haskell> |
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| − | binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y)) |
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| − | prodxy x y=product[x..y] |
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| − | problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]] |
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| − | </haskell> |
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| − | == [http://projecteuler.net/index.php?section=problems&id=114 Problem 114] == |
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| − | Investigating the number of ways to fill a row with separated blocks that are at least three units long. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | -- fun in p115 |
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| − | problem_114=fun 3 50 |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=115 Problem 115] == |
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| − | Finding a generalisation for the number of ways to fill a row with separated blocks. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y)) |
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| − | prodxy x y=product[x..y] |
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| − | fun m n=sum[binomial (k+a) (k-a)|a<-[0..div (n+1) (m+1)],let k=1-a*m+n] |
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| − | problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]] |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=116 Problem 116] == |
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| − | Investigating the number of ways of replacing square tiles with one of three coloured tiles. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y)) |
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| − | prodxy x y=product[x..y] |
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| − | f116 n x=sum[binomial (a+b) a|a<-[1..div n x],let b=n-a*x] |
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| − | p116 x=sum[f116 x a|a<-[2..4]] |
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| − | problem_116 = p116 50 |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=117 Problem 117] == |
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| − | Investigating the number of ways of tiling a row using different-sized tiles. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | fibs5 = 0 : 0 :1: 1:zipWith4 (\a b c d->a+b+c+d) fibs5 a1 a2 a3 |
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| − | where |
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| − | a1=tail fibs5 |
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| − | a2=tail a1 |
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| − | a3=tail a2 |
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| − | p117 x=fibs5!!(x+2) |
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| − | problem_117 = p117 50 |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=118 Problem 118] == |
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| − | Exploring the number of ways in which sets containing prime elements can be made. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | digits = ['1'..'9'] |
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| − | |||
| − | -- possible partitions voor prime number sets |
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| − | -- leave out patitions with more than 4 1's |
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| − | -- because only {2,3,5,7,..} is possible |
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| − | -- and the [9]-partition because every permutation of all |
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| − | -- nine digits is divisable by 3 |
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| − | test xs |
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| − | |len>4=False |
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| − | |xs==[9]=False |
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| − | |otherwise=True |
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| − | where |
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| − | len=length $filter (==1) xs |
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| − | parts = filter test $partitions 9 |
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| − | permutationsOf [] = [[]] |
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| − | permutationsOf xs = [x:xs' | x <- xs, xs' <- permutationsOf (delete x xs)] |
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| − | combinationsOf 0 _ = [[]] |
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| − | combinationsOf _ [] = [] |
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| − | combinationsOf k (x:xs) = |
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| − | map (x:) (combinationsOf (k-1) xs) ++ combinationsOf k xs |
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| − | |||
| − | priemPerms [] = 0 |
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| − | priemPerms ds = |
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| − | fromIntegral . length . filter (isPrime . read) . permutationsOf $ ds |
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| − | setsums [] 0 = [[]] |
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| − | setsums [] _ = [] |
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| − | setsums (x:xs) n |
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| − | | x > n = setsums xs n |
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| − | | otherwise = map (x:) (setsums (x:xs) (n-x)) ++ setsums xs n |
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| − | |||
| − | partitions n = setsums (reverse [1..n]) n |
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| − | |||
| − | fc :: [Integer] -> [Char] -> Integer |
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| − | fc (p:[]) ds = priemPerms ds |
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| − | fc (p:ps) ds = |
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| − | foldl fcmul 0 . combinationsOf p $ ds |
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| − | where |
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| − | fcmul x y |
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| − | | np y == 0 = x |
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| − | | otherwise = x + np y * fc ps (ds \\ y) |
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| − | where |
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| − | np = priemPerms |
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| − | -- here is the 'imperfection' correction method: |
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| − | -- make use of duplicate reducing factors for partitions |
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| − | -- with repeating factors, f.i. [1,1,1,1,2,3]: |
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| − | -- in this case 4 1's -> factor = 4! |
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| − | -- or for [1,1,1,3,3] : factor = 3! * 2! |
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| − | dupF :: [Integer] -> Integer |
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| − | dupF = foldl (\ x y -> x * product [1..y]) 1 . map (fromIntegral . length) . group |
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| − | |||
| − | main = do |
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| − | print . sum . map (\x -> fc x digits `div` dupF x) $ parts |
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| − | problem_118 = main |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=119 Problem 119] == |
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| − | Investigating the numbers which are equal to sum of their digits raised to some power. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | import Data.List |
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| − | digits n |
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| − | {- 123->[3,2,1] |
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| − | -} |
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| − | |n<10=[n] |
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| − | |otherwise= y:digits x |
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| − | where |
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| − | (x,y)=divMod n 10 |
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| − | problem_119 =sort [(a^b)| |
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| − | a<-[2..200], |
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| − | b<-[2..9], |
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| − | let m=a^b, |
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| − | let n=sum$digits m, |
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| − | n==a]!!29 |
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| − | </haskell> |
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| − | |||
| − | == [http://projecteuler.net/index.php?section=problems&id=120 Problem 120] == |
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| − | Finding the maximum remainder when (a − 1)n + (a + 1)n is divided by a2. |
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| − | |||
| − | Solution: |
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| − | <haskell> |
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| − | fun m=div (m*(8*m^2-3*m-5)) 3 |
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| − | problem_120 = fun 500 |
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| − | </haskell> |
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Revision as of 21:45, 29 January 2008
Do them on your own!