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Euler problems/111 to 120

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Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_120 = undefined
+
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]]
 
</haskell>
 
</haskell>

Revision as of 11:42, 13 December 2007

Contents

1 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"

2 Problem 112

Investigating the density of "bouncy" numbers.

Solution:

import Data.List
digits n 
{-  change 123 to [3,2,1]
 -}
    |n<10=[n]
    |otherwise= y:digits x 
    where
    (x,y)=divMod n 10
isdecr x=
    null$filter (\(x, y)->x-y<0)$zip di k
    where
    di=digits x
    k=0:di
isincr x=
    null$filter (\(x, y)->x-y<0)$zip di k
    where
    di=digits x
    k=tail$di++[0]
nnn=1500000
num150 =length [x|x<-[1..nnn],isdecr x||isincr x]
istwo x|isdecr x||isincr x=1
     |otherwise=0
problem_112 n1 n2=
    if (div n1 n2==100)
       then do appendFile "file.log" ((show n1)  ++"   "++ (show n2)++"\n")
               return()
       else  problem_112 (n1+1) (n2+istwo (n1+1))
main=  problem_112 nnn num150

3 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:

import List
series  2 =replicate 10 1
series n=sumkey$map (\(x, y)->map (*y) x)$zip key (series (n-1))
key =[replicate (a+1) 1++replicate (9-a) 0|a<-[0..9]]
sumkey k=[sum [a!!m|a<-k]|m<-[0..9]]
fun x= sum [(sum$series i)-1|i<-[2..x]]-(x-1)*9-1+(sum$series  x)
problem_113 =fun 101

4 Problem 114

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

Solution:

problem_114 = undefined

5 Problem 115

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

Solution:

problem_115 = undefined

6 Problem 116

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

Solution:

problem_116 = undefined

7 Problem 117

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

Solution:

problem_117 = undefined

8 Problem 118

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

Solution:

problem_118 = undefined

9 Problem 119

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

Solution:

problem_119 = undefined

10 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]]