Difference between revisions of "Euler problems/151 to 160"

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== [http://projecteuler.net/index.php?section=problems&id=151 Problem 151] ==
+
Do them on your own!
Paper sheets of standard sizes: an expected-value problem.
 
 
 
Solution:
 
<haskell>
 
problem_151 = undefined
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=152 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:
 
<haskell>
 
import Data.Ratio
 
import Data.List
 
 
 
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]
 
 
 
-- Verify that we need not consider terms divisible by 11, or by any
 
-- prime greater than 13. Nor need we consider any term divisible
 
-- by 25, 27, 32, or 49.
 
verify = all (\p -> null $ tail $ pfree [p, 2*p..85] 0 p) $
 
        11 : dropWhile (< 17) primes ++ [25, 27, 32, 49]
 
 
 
-- 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 f [(0, [[]])] [13, 7, 5]
 
  where
 
    f 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  (x, _) (y, _) = x == y
 
    cmpFst (x, _) (y, _) = compare x y
 
 
 
-- All subsets (of an ordered set) whose sum of inverse squares is x
 
findInvSq x y = f x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y)
 
  where
 
    f 0 _        = [[]]
 
    f x ((n, r, s):ns)
 
    | r > x    = f x ns
 
    | s < x    = []
 
    | otherwise = map (n :) (f (x - r) ns) ++ f x ns
 
    f _ _        = []
 
 
 
-- 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 = if verify
 
              then [sort $ u ++ v | (x, s) <- only23,
 
                                    u <- findInvSq (1%2 - x) all23,
 
                                    v <- s]
 
              else undefined
 
 
 
problem_152 = length solutions
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=153 Problem 153] ==
 
Investigating Gaussian Integers
 
 
 
Solution:
 
<haskell>
 
problem_153 = undefined
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=154 Problem 154] ==
 
Exploring Pascal's pyramid.
 
 
 
Solution:
 
<haskell>
 
#include <stdio.h>
 
int main(){
 
    int bound = 200000;
 
    long long sum = 0;
 
    int val2[bound+1], val5[bound+1]; // number of factors 2/5 in i!
 
    int v2 = 0, v5  = 0;
 
    int i;
 
    int n;
 
    for(n=0;n<=bound;n++)
 
    {val5[n]=n/5+n/25+n/125+n/625+n/3125+n/15625+n/78125;
 
        val2[n]=n/2+n/4+n/8+n/16+n/32+n/64+n/128+n/256+n/512+n/1024
 
            +n/2048+n/4096+n/8192+n/16384+n/32768+n/65536+n/131072;}
 
 
 
        v2 =val2[bound]- 11;
 
        v5 = val5[bound]-11;
 
        int j,k,vi2,vi5;
 
        for(i = 2; i < 65625; i++){
 
            if (!(i&1023)){
 
                // look how many we got so far
 
                printf("%d:\t%lld\n",i,sum);
 
            }
 
            vi5 = val5[i];
 
            vi2 = val2[i];
 
            int jb = ((bound - i) >> 1)+1;
 
            // I want i <= j <= k
 
            // by carry analysis, I know that if i < 4*5^5+2, then
 
            // j must be at least 2*5^6+2
 
            for(j = (i < 12502) ? 31252 : i; j < jb; j++){
 
                k = bound - i - j;
 
                if (vi5 + val5[j] + val5[k] < v5
 
                        && vi2 + val2[j] + val2[k] < v2){
 
                    if (j == k || i == j){
 
                        sum += 3;
 
                    } else {
 
                        sum += 6;
 
                    }
 
                }
 
            }
 
        }
 
        printf("Total:\t%lld\n",sum);
 
        return 0;
 
}
 
problem_154 = main
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=155 Problem 155] ==
 
Counting Capacitor Circuits.
 
 
 
Solution:
 
<haskell>
 
problem_155 = undefined
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=156 Problem 156] ==
 
Counting Digits
 
 
 
Solution:
 
<haskell>
 
digits =reverse.digits'
 
    where
 
    digits' n
 
        |n<10=[n]
 
        |otherwise= y:digits' x
 
        where
 
        (x,y)=divMod n 10
 
digitsToNum n=foldl dmm 0  n
 
    where
 
    dmm=(\x y->x*10+y)
 
countA :: Int -> Integer
 
countA 0 = 0
 
countA k = fromIntegral k * (10^(k-1))
 
 
countFun :: Integer -> Integer -> Integer
 
countFun _ 0 = 0
 
countFun d n = countL ds k
 
      where
 
        ds = digits n
 
        k = length ds - 1
 
        countL [a] _
 
            | a < d    = 0
 
            | otherwise = 1
 
        countL (a:tl) m
 
            | a < d    = a*countA m + countL tl (m-1)
 
            | a == d    = a*countA m + digitsToNum tl + 1 + countL tl (m-1)
 
            | otherwise = a*countA m + 10^m + countL tl (m-1)
 
 
fixedPoints :: Integer -> [Integer]
 
fixedPoints d
 
    = [a*10^10+b | a <- [0 .. d-1], b <- findFrom 0 (10^10-1)]
 
      where
 
        fun = countFun d
 
        good r = r == fun r
 
        findFrom lo hi
 
            | hi < lo  = []
 
            | good lo  = lgs ++ findFrom (last lgs + 2) hi
 
            | good hi  = findFrom lo (last hgs - 2) ++ reverse hgs
 
            | h1 < l1  = []
 
            | l1 == h1  = if good l1 then [l1] else []
 
            | m0 == m1  = findFrom l1 (head mgs - 2) ++ mgs
 
                            ++ findFrom (last mgs + 2) h1
 
            | m0 < m1  = findFrom l1 (m0-1) ++ findFrom (goUp h1 m1) h1
 
            | otherwise = findFrom l1 (goDown l1 m1) ++ findFrom (m0+1) h1
 
              where
 
                l1 = goUp hi lo
 
                h1 = goDown l1 hi
 
                goUp bd k
 
                    | k < k1 && k < bd  = goUp bd k1
 
                    | otherwise        = k
 
                      where
 
                        k1 = fun k
 
                goDown bd k
 
                    | k1 < k && bd < k  = goDown bd k1
 
                    | otherwise        = k
 
                      where
 
                        k1 = fun k
 
                m0 = (l1 + h1) `div` 2
 
                m1 = fun m0
 
                lgs = takeWhile good [lo .. hi]
 
                hgs = takeWhile good [hi,hi-1 .. lo]
 
                mgs = reverse (takeWhile good [m0,m0-1 .. l1])
 
                        ++ takeWhile good [m0+1 .. h1]
 
problem_156=sum[sum $fixedPoints a|a<-[1..9]]
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=157 Problem 157] ==
 
Solving the diophantine equation 1/a+1/b= p/10n
 
 
 
Solution:
 
<haskell>
 
problem_157 = undefined
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=158 Problem 158] ==
 
Exploring strings for which only one character comes lexicographically after its neighbour to the left.
 
 
 
Solution:
 
<haskell>
 
problem_158 = undefined
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=159 Problem 159] ==
 
Digital root sums of factorisations.
 
 
 
Solution:
 
<haskell>
 
import Control.Monad
 
import Data.Array.ST
 
import qualified Data.Array.Unboxed as U
 
spfArray :: U.UArray Int Int
 
spfArray  = runSTUArray (do
 
  arr <- newArray (0,m-1) 0
 
  loop arr 2
 
  forM_ [2 .. m - 1] $ \ x ->
 
    loop2 arr x 2
 
  return arr
 
  )
 
  where
 
  m=10^6
 
  loop arr n
 
      |n>=m=return ()
 
      |otherwise=do writeArray arr n (n-9*(div (n-1) 9))
 
                    loop arr (n+1)
 
  loop2 arr x n
 
      |n*x>=m=return ()
 
      |otherwise=do incArray arr x n
 
                    loop2 arr x (n+1)
 
  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"$foldl (++) "" [show x,"\n"]
 
main=do
 
    mapM_ writ $U.elems spfArray
 
problem_159 = main
 
 
 
--at first ,make main to get file "p159.log"
 
--then ,add all num in the file
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=160 Problem 160] ==
 
Factorial trailing digits
 
 
 
We use the following two facts:
 
 
 
Fact 1: <hask>(2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0</hask>
 
 
 
Fact 2: <hask>product [n | n <- [0..10^d], gcd n 10 == 1] `mod` 10^d == 1</hask>
 
 
 
We really only need these two facts for the special case of
 
<hask>d == 5</hask>, 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 <hask>5^d</hask> has
 
<hask>4*5^(d-1)</hask> elements.
 
 
 
Fact 2 follows from the fact that the group of invertible elements
 
of the ring of integers modulo <hask>10^d</hask> is isomorphic to the product
 
of a cyclic group of order 2 and another cyclic group.
 
 
 
Solution:
 
<haskell>
 
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)
 
</haskell>
 
it have another fast way to do this .
 
 
 
Solution:
 
<haskell>
 
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=mod (a*hFacial(a-1)) (5^5)
 
    |otherwise=hFacial(a-1)
 
fastFacial a= hFacial $mod a 6250
 
numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]]
 
p160 x=mulMod t5 a b
 
    where
 
    t5=10^5
 
    lst=numPrime x 5
 
    a=powMod t5 1563 $mod c 2500
 
    b=productMod  c6
 
    c=sum lst
 
    c6=map fastFacial $x:lst
 
problem_160 = p160 (10^12)
 
 
 
</haskell>
 

Revision as of 21:43, 29 January 2008

Do them on your own!