Euler problems/151 to 160
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Problem 151
Paper sheets of standard sizes: an expected-value problem.
Solution:
problem_151 = fun (1,1,1,1)
fun (0,0,0,1) = 0
fun (0,0,1,0) = fun (0,0,0,1) + 1
fun (0,1,0,0) = fun (0,0,1,1) + 1
fun (1,0,0,0) = fun (0,1,1,1) + 1
fun (a,b,c,d) =
(pickA + pickB + pickC + pickD) / (a + b + c + d)
where
pickA | a > 0 = a * fun (a-1,b+1,c+1,d+1)
| otherwise = 0
pickB | b > 0 = b * fun (a,b-1,c+1,d+1)
| otherwise = 0
pickC | c > 0 = c * fun (a,b,c-1,d+1)
| otherwise = 0
pickD | d > 0 = d * fun (a,b,c,d-1)
| otherwise = 0
Problem 152
Writing 1/2 as a sum of inverse squares
Note that if p is an odd prime, the sum of inverse squares of all terms divisible by p must have reduced denominator not divisible by p.
Solution:
import Data.Ratio
import Data.List
import Data.Ord (comparing)
import Data.Function (on)
invSq n = 1 % (n * n)
sumInvSq = sum . map invSq
subsets (x:xs) = let s = subsets xs in s ++ map (x :) s
subsets _ = [[]]
primes = 2 : 3 : 7 : [p | p <- [11, 13..79],
all (\q -> p `mod` q /= 0) [3, 5, 7]]
-- All subsets whose sum of inverse squares,
-- when added to x, does not contain a factor of p
pfree s x p = [(y, t) | t <- subsets s, let y = x + sumInvSq t,
denominator y `mod` p /= 0]
-- All pairs (x, s) where x is a rational number whose reduced
-- denominator is not divisible by any prime greater than 3;
-- and s is all sets of numbers up to 80 divisible
-- by a prime greater than 3, whose sum of inverse squares is x.
only23 = foldl fun [(0, [[]])] [13, 7, 5]
where
fun a p =
collect $ [(y, u ++ v) |
(x, s) <- a,
(y, v) <- pfree (terms p) x p,
u <- s]
terms p =
[n * p |
n <- [1..80`div`p],
all (\q -> n `mod` q /= 0) $
11 : takeWhile (>= p) [13, 7, 5]
]
collect =
map (\z -> (fst $ head z, map snd z)) .
groupBy fstEq . sortBy cmpFst
fstEq = (==) `on` fst
cmpFst = comparing fst
-- All subsets (of an ordered set) whose sum of inverse squares is x
findInvSq x y =
fun x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y)
where
fun 0 _ = [[]]
fun x ((n, r, s):ns)
| r > x = fun x ns
| s < x = []
| otherwise = map (n :) (fun (x - r) ns) ++ fun x ns
fun _ _ = []
-- All numbers up to 80 that are divisible only by the primes
-- 2 and 3 and are not divisible by 32 or 27.
all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 80]
solutions =
[sort $ u ++ v |
(x, s) <- only23,
u <- findInvSq (1%2 - x) all23,
v <- s
]
problem_152 = length solutions
Problem 153
Investigating Gaussian Integers
Problem 154
Exploring Pascal's pyramid.
Problem 155
Counting Capacitor Circuits.
Solution:
--http://www.research.att.com/~njas/sequences/A051389
a051389=
[1, 2, 4, 8, 20, 42,
102, 250, 610, 1486,
3710, 9228, 23050, 57718,
145288, 365820, 922194, 2327914
]
problem_155 = sum a051389
Problem 156
Counting Digits
Problem 157
Solving the diophantine equation 1/a+1/b= p/10n
Solution:
-- Call (a,b,p) a primitive tuple of equation 1/a+1/b=p/10^n
-- a and b are divisors of 10^n, gcd a b == 1, a <= b and a*b <= 10^n
-- I noticed that the number of variants with a primitive tuple
-- is equal to the number of divisors of p.
-- So I produced all possible primitive tuples per 10^n and
-- summed all the number of divisors of every p
import Data.List
k `divides` n = n `mod` k == 0
divisors n
| n == 10 = [1,2,5,10]
| otherwise =
[ d |
d <- [1..n `div` 5],
d `divides` n ]
++ [n `div` 4, n `div` 2,n]
fp n =
[ n*(a+b) `div` ab |
a <- ds,
b <- dropWhile (<a) ds,
gcd a b == 1,
let ab = a*b,
ab <= n
]
where
ds = divisors n
numDivisors :: Integer -> Integer
numDivisors n = product [ toInteger (a+1) | (p,a) <- primePowerFactors n]
numVgln = sum . map numDivisors . fp
main = do
print . sum . map numVgln . takeWhile (<=10^9) . iterate (10*) $ 10
primePowerFactors x = [(head a ,length a)|a<-group$primeFactors x]
merge xs@(x:xt) ys@(y:yt) = case compare x y of
LT -> x : (merge xt ys)
EQ -> x : (merge xt yt)
GT -> y : (merge xs yt)
diff xs@(x:xt) ys@(y:yt) = case compare x y of
LT -> x : (diff xt ys)
EQ -> diff xt yt
GT -> diff xs yt
primes, nonprimes :: [Integer]
primes = [2,3,5] ++ (diff [7,9..] nonprimes)
nonprimes = foldr1 f . map g $ tail primes
where f (x:xt) ys = x : (merge xt ys)
g p = [ n*p | n <- [p,p+2..]]
primeFactors n =
factor n primes
where
factor n (p:ps)
| p*p > n = [n]
| n `mod` p == 0 = p : factor (n `div` p) (p:ps)
| otherwise = factor n ps
Problem 158
Exploring strings for which only one character comes lexicographically after its neighbour to the left.
Solution:
factorial n = product [1..toInteger n]
fallingFactorial x n = product [x - i | i <- [0..fromIntegral n - 1] ]
choose n k = fallingFactorial n k `div` factorial k
fun n=(2 ^ n - n - 1) * choose 26 n
problem_158=maximum$map fun [1..26]
Problem 159
Digital root sums of factorisations.
Solution:
import Control.Monad
import Data.Array.ST
import qualified Data.Array.Unboxed as U
spfArray :: U.UArray Int Int
spfArray = runSTUArray (do
arr <- newArray (2,m-1) 0
forM_ [2 .. m-1] $ \n ->
writeArray arr n (n-9*((n-1) `div` 9))
forM_ [2 .. m-1] $ \x ->
forM_ [2 .. m`div`n-1] $ \n ->
incArray arr x n
return arr
)
where
m=10^6
incArray arr x n = do
a <- readArray arr x
b <- readArray arr n
ab <- readArray arr (x*n)
when(ab<a+b) (writeArray arr (x*n) (a + b))
writ x=appendFile "p159.log"$ show x ++ "\n"
main=mapM_ writ $U.elems spfArray
problem_159 = main
--at first ,make main to get file "p159.log"
--then ,add all num in the file
Problem 160
Factorial trailing digits
We use the following two facts:
Fact 1: (2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0
Fact 2: product [n | n <- [0..10^d], gcd n 10 == 1] `mod` 10^d == 1
We really only need these two facts for the special case of
d == 5, and we can verify that directly by
evaluating the above two Haskell expressions.
More generally:
Fact 1 follows from the fact that the group of invertible elements
of the ring of integers modulo 5^d has
4*5^(d-1) elements.
Fact 2 follows from the fact that the group of invertible elements
of the ring of integers modulo 10^d is isomorphic to the product
of a cyclic group of order 2 and another cyclic group.
Solution:
problem_160 = trailingFactorialDigits 5 (10^12)
trailingFactorialDigits d n = twos `times` odds
where
base = 10 ^ d
x `times` y = (x * y) `mod` base
multiply = foldl' times 1
x `toPower` k = multiply $ genericReplicate n x
e = facFactors 2 n - facFactors 5 n
twos
| e <= d = 2 `toPower` e
| otherwise = 2 `toPower` (d + (e - d) `mod` (4 * 5 ^ (d - 1)))
odds = multiply [odd | a <- takeWhile (<= n) $ iterate (* 2) 1,
b <- takeWhile (<= n) $ iterate (* 5) a,
odd <- [3, 5 .. n `div` b `mod` base],
odd `mod` 5 /= 0]
-- The number of factors of the prime p in n!
facFactors p = sum . zipWith (*) (iterate (\x -> p * x + 1) 1) .
tail . radix p
-- The digits of n in base b representation
radix p = map snd . takeWhile (/= (0, 0)) .
iterate ((`divMod` p) . fst) . (`divMod` p)
it have another fast way to do this .
Solution:
import Data.List
mulMod :: Integral a => a -> a -> a -> a
mulMod a b c= (b * c) `rem` a
squareMod :: Integral a => a -> a -> a
squareMod a b = (b * b) `rem` a
pow' :: (Num a, Integral b) => (a -> a -> a) -> (a -> a) -> a -> b -> a
pow' _ _ _ 0 = 1
pow' mul sq x' n' = f x' n' 1
where
f x n y
| n == 1 = x `mul` y
| r == 0 = f x2 q y
| otherwise = f x2 q (x `mul` y)
where
(q,r) = quotRem n 2
x2 = sq x
powMod :: Integral a => a -> a -> a -> a
powMod m = pow' (mulMod m) (squareMod m)
productMod =foldl (mulMod (10^5)) 1
hFacial 0=1
hFacial a
|gcd a 5==1=(a*hFacial(a-1)) `mod` (5^5)
|otherwise=hFacial(a-1)
fastFacial a= hFacial $a `mod` 6250
numPrime x p=takeWhile(>0) [x `div` (p^a)|a<-[1..]]
p160 x=mulMod t5 a b
where
t5=10^5
lst=numPrime x 5
a=powMod t5 1563 $c `mod` 2500
b=productMod c6
c=sum lst
c6=map fastFacial $x:lst
problem_160 = p160 (10^12)