Difference between revisions of "Euler problems/111 to 120"
Line 151: | Line 151: | ||
Solution: |
Solution: |
||
<haskell> |
<haskell> |
||
+ | import List |
||
− | problem_120 = undefined |
||
+ | 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
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:
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
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$series x)
problem_113 =fun 101
Problem 114
Investigating the number of ways to fill a row with separated blocks that are at least three units long.
Solution:
problem_114 = undefined
Problem 115
Finding a generalisation for the number of ways to fill a row with separated blocks.
Solution:
problem_115 = undefined
Problem 116
Investigating the number of ways of replacing square tiles with one of three coloured tiles.
Solution:
problem_116 = undefined
Problem 117
Investigating the number of ways of tiling a row using different-sized tiles.
Solution:
problem_117 = undefined
Problem 118
Exploring the number of ways in which sets containing prime elements can be made.
Solution:
problem_118 = undefined
Problem 119
Investigating the numbers which are equal to sum of their digits raised to some power.
Solution:
problem_119 = undefined
Problem 120
Finding the maximum remainder when (a − 1)n + (a + 1)n is divided by a2.
Solution:
import List
primes :: [Integer]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
primeFactors :: Integer -> [Integer]
primeFactors n = factor n primes
where
factor _ [] = []
factor m (p:ps) | p*p > m = [m]
| m `mod` p == 0 = p : factor (m `div` p) (p:ps)
| otherwise = factor m ps
isPrime :: Integer -> Bool
isPrime 1 = False
isPrime n = case (primeFactors n) of
(_:_:_) -> False
_ -> True
fun x
|even x=x*(x-2)
|not$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]]