Euler problems/111 to 120

Problem 111

Search for 10-digit primes containing the maximum number of repeated digits.

Solution:

import Control.Monad (replicateM)

-- All ways of interspersing n copies of x into a list
intr :: Int -> a -> [a] -> [[a]]
intr 0 _ y      = [y]
intr n x (y:ys) = concat
                  [map ((replicate i x ++) . (y :)) $ intr (n-i) x ys
                       | i <- [0..n]]
intr n x _      = [replicate n x]

-- All 10-digit primes containing the maximal number of the digit d
maxDigits :: Char -> [Integer]
maxDigits d = head $ dropWhile null
              [filter isPrime $ map read $ filter ((/='0') . head) $
               concatMap (intr (10-n) d) $
               replicateM n $ delete d "0123456789"
                   | n <- [1..9]]
 
problem_111 = sum $ concatMap maxDigits "0123456789"

Problem 112

Investigating the density of "bouncy" numbers.

Solution:

isIncreasing' n p
    | n == 0 = True
    | p >= p1 = isIncreasing' (n `div` 10) p1
    | otherwise = False
    where
    p1 = n `mod` 10

isIncreasing :: Int -> Bool
isIncreasing n = isIncreasing' (n `div` 10) (n `mod` 10)

isDecreasing' n p
    | n == 0 = True
    | p <= p1 = isDecreasing' (n `div` 10) p1
    | otherwise = False
    where
    p1 = n `mod` 10

isDecreasing :: Int -> Bool
isDecreasing n = isDecreasing' (n `div` 10) (n `mod` 10)

isBouncy n = not (isIncreasing n) && not (isDecreasing n)
nnn=1500000
num150 =length [x|x<-[1..nnn],isBouncy x]
p112 n nb
    | fromIntegral nnb / fromIntegral n >= 0.99 = n
    | otherwise = prob112' (n+1) nnb
    where 
    nnb = if isBouncy n then nb + 1 else nb

problem_112=p112 (nnn+1) num150

Problem 113

How many numbers below a googol (10100) are not "bouncy"?

Solution:

import Array

mkArray b f = listArray b $ map f (range b)

digits = 100

inc = mkArray ((1, 0), (digits, 9)) ninc
dec = mkArray ((1, 0), (digits, 9)) ndec

ninc (1, _) = 1
ninc (l, d) = sum [inc ! (l-1, i) | i <- [d..9]]

ndec (1, _) = 1
ndec (l, d) = sum [dec ! (l-1, i) | i <- [0..d]]

problem_113 = sum [inc ! i | i <- range ((digits, 0), (digits, 9))]
               + sum [dec ! i | i <- range ((1, 1), (digits, 9))]
               - digits*9 -- numbers like 11111 are counted in both inc and dec 
               - 1 -- 0 is included in the increasing numbers

Note: inc and dec contain the same data, but it seems clearer to duplicate them.

it is another way to solution this problem:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
prodxy x y=product[x..y]
problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]]

Problem 114

Investigating the number of ways to fill a row with separated blocks that are at least three units long.

Solution:

-- fun in p115
problem_114=fun 3 50

Problem 115

Finding a generalisation for the number of ways to fill a row with separated blocks.

Solution:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
prodxy x y=product[x..y]
fun m n=sum[binomial (k+a) (k-a)|a<-[0..div (n+1) (m+1)],let k=1-a*m+n]
problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]]

Problem 116

Investigating the number of ways of replacing square tiles with one of three coloured tiles.

Solution:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
prodxy x y=product[x..y]
f116 n x=sum[binomial (a+b) a|a<-[1..div n x],let b=n-a*x]
p116 x=sum[f116 x a|a<-[2..4]]
problem_116 = p116 50

Problem 117

Investigating the number of ways of tiling a row using different-sized tiles.

Solution:

fibs5 = 0 : 0 :1: 1:zipWith4 (\a b c d->a+b+c+d) fibs5 a1 a2 a3 
    where
    a1=tail fibs5
    a2=tail a1
    a3=tail a2
p117 x=fibs5!!(x+2)
problem_117 = p117 50

Problem 118

Exploring the number of ways in which sets containing prime elements can be made.

Solution:

problem_118 = undefined

Problem 119

Investigating the numbers which are equal to sum of their digits raised to some power.

Solution:

import Data.List
digits n 
{-  123->[3,2,1]
 -}
    |n<10=[n]
    |otherwise= y:digits x 
    where
    (x,y)=divMod n 10
problem_119 =sort [(a^b)|
    a<-[2..200],
    b<-[2..9],
    let m=a^b,
    let n=sum$digits m,
    n==a]!!29

Problem 120

Finding the maximum remainder when (a − 1)n + (a + 1)n is divided by a2.

Solution:

import List
primes :: [Integer]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 
primeFactors :: Integer -> [Integer]
primeFactors n = factor n primes
    where
        factor _ [] = []
        factor m (p:ps) | p*p > m        = [m]
                        | m `mod` p == 0 = p : factor (m `div` p) (p:ps)
                        | otherwise      = factor m ps
 
isPrime :: Integer -> Bool
isPrime 1 = False
isPrime n = case (primeFactors n) of
                (_:_:_)   -> False
                _         -> True
fun x
    |even x=x*(x-2)
    |not$null$funb x=head$funb x
    |odd e=x*(x-1)
    |otherwise=2*x*(e-1)
    where
    e=div x 2

funb x=take 1 [nn*x|
    a<-[1,3..x],
    let n=div (x-1) 2,
    let p=x*a+n,
    isPrime p,
    let nn=mod (2*(x*a+n)) x
    ]

problem_120 = sum [fun a|a<-[3..1000]]