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Euler problems/151 to 160

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== [http://projecteuler.net/index.php?section=problems&id=151 Problem 151] ==
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Do them on your own!
Paper sheets of standard sizes: an expected-value problem.
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Solution:
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<haskell>
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problem_151 = undefined
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=152 Problem 152] ==
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Writing 1/2 as a sum of inverse squares
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Note that if p is an odd prime, the sum of inverse squares of
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all terms divisible by p must have reduced denominator not divisible
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by p.
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Solution:
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<haskell>
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import Data.Ratio
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import Data.List
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invSq n = 1 % (n * n)
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sumInvSq = sum . map invSq
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subsets (x:xs) = let s = subsets xs in s ++ map (x :) s
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subsets _      = [[]]
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primes = 2 : 3 : 7 : [p | p <- [11, 13..79],
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                          all (\q -> p `mod` q /= 0) [3, 5, 7]]
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-- All subsets whose sum of inverse squares,
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-- when added to x, does not contain a factor of p
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pfree s x p = [(y, t) | t <- subsets s, let y =  x + sumInvSq t,
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                        denominator y `mod` p /= 0]
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-- Verify that we need not consider terms divisible by 11, or by any
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-- prime greater than 13. Nor need we consider any term divisible
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-- by 25, 27, 32, or 49.
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verify = all (\p -> null $ tail $ pfree [p, 2*p..85] 0 p) $
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        11 : dropWhile (< 17) primes ++ [25, 27, 32, 49]
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-- All pairs (x, s) where x is a rational number whose reduced
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-- denominator is not divisible by any prime greater than 3;
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-- and s is all sets of numbers up to 80 divisible
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-- by a prime greater than 3, whose sum of inverse squares is x.
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only23 = foldl f [(0, [[]])] [13, 7, 5]
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  where
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    f a p = collect $ [(y, u ++ v) | (x, s) <- a,
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                                    (y, v) <- pfree (terms p) x p,
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                                    u <- s]
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    terms p = [n * p | n <- [1..80`div`p],
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                      all (\q -> n `mod` q /= 0) $
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                          11 : takeWhile (>= p) [13, 7, 5]
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              ]
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    collect = map (\z -> (fst $ head z, map snd z)) .
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              groupBy fstEq . sortBy cmpFst
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    fstEq  (x, _) (y, _) = x == y
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    cmpFst (x, _) (y, _) = compare x y
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-- All subsets (of an ordered set) whose sum of inverse squares is x
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findInvSq x y = f x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y)
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  where
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    f 0 _        = [[]]
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    f x ((n, r, s):ns)
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    | r > x    = f x ns
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    | s < x    = []
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    | otherwise = map (n :) (f (x - r) ns) ++ f x ns
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    f _ _        = []
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-- All numbers up to 80 that are divisible only by the primes
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-- 2 and 3 and are not divisible by 32 or 27.
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all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 80]
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solutions = if verify
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              then [sort $ u ++ v | (x, s) <- only23,
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                                    u <- findInvSq (1%2 - x) all23,
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                                    v <- s]
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              else undefined
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problem_152 = length solutions
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=153 Problem 153] ==
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Investigating Gaussian Integers
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Solution:
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<haskell>
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problem_153 = undefined
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=154 Problem 154] ==
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Exploring Pascal's pyramid.
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Solution:
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<haskell>
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#include <stdio.h>
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int main(){
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    int bound = 200000;
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    long long sum = 0;
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    int val2[bound+1], val5[bound+1]; // number of factors 2/5 in i!
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    int v2 = 0, v5  = 0;
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    int i;
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    int n;
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    for(n=0;n<=bound;n++)
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    {val5[n]=n/5+n/25+n/125+n/625+n/3125+n/15625+n/78125;
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        val2[n]=n/2+n/4+n/8+n/16+n/32+n/64+n/128+n/256+n/512+n/1024
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            +n/2048+n/4096+n/8192+n/16384+n/32768+n/65536+n/131072;}
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        v2 =val2[bound]- 11;
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        v5 = val5[bound]-11;
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        int j,k,vi2,vi5;
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        for(i = 2; i < 65625; i++){
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            if (!(i&1023)){
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                // look how many we got so far
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                printf("%d:\t%lld\n",i,sum);
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            }
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            vi5 = val5[i];
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            vi2 = val2[i];
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            int jb = ((bound - i) >> 1)+1;
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            // I want i <= j <= k
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            // by carry analysis, I know that if i < 4*5^5+2, then
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            // j must be at least 2*5^6+2
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            for(j = (i < 12502) ? 31252 : i; j < jb; j++){
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                k = bound - i - j;
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                if (vi5 + val5[j] + val5[k] < v5
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                        && vi2 + val2[j] + val2[k] < v2){
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                    if (j == k || i == j){
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                        sum += 3;
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                    } else {
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                        sum += 6;
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                    }
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                }
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            }
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        }
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        printf("Total:\t%lld\n",sum);
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        return 0;
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}
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problem_154 = main
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=155 Problem 155] ==
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Counting Capacitor Circuits.
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Solution:
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<haskell>
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problem_155 = undefined
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=156 Problem 156] ==
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Counting Digits
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Solution:
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<haskell>
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digits =reverse.digits'
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    where
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    digits' n
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        |n<10=[n]
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        |otherwise= y:digits' x
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        where
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        (x,y)=divMod n 10
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digitsToNum n=foldl dmm 0  n
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    where
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    dmm=(\x y->x*10+y)
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countA :: Int -> Integer
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countA 0 = 0
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countA k = fromIntegral k * (10^(k-1))
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countFun :: Integer -> Integer -> Integer
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countFun _ 0 = 0
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countFun d n = countL ds k
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      where
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        ds = digits n
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        k = length ds - 1
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        countL [a] _
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            | a < d    = 0
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            | otherwise = 1
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        countL (a:tl) m
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            | a < d    = a*countA m + countL tl (m-1)
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            | a == d    = a*countA m + digitsToNum tl + 1 + countL tl (m-1)
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            | otherwise = a*countA m + 10^m + countL tl (m-1)
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fixedPoints :: Integer -> [Integer]
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fixedPoints d
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    = [a*10^10+b | a <- [0 .. d-1], b <- findFrom 0 (10^10-1)]
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      where
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        fun = countFun d
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        good r = r == fun r
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        findFrom lo hi
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            | hi < lo  = []
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            | good lo  = lgs ++ findFrom (last lgs + 2) hi
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            | good hi  = findFrom lo (last hgs - 2) ++ reverse hgs
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            | h1 < l1  = []
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            | l1 == h1  = if good l1 then [l1] else []
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            | m0 == m1  = findFrom l1 (head mgs - 2) ++ mgs
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                            ++ findFrom (last mgs + 2) h1
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            | m0 < m1  = findFrom l1 (m0-1) ++ findFrom (goUp h1 m1) h1
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            | otherwise = findFrom l1 (goDown l1 m1) ++ findFrom (m0+1) h1
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              where
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                l1 = goUp hi lo
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                h1 = goDown l1 hi
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                goUp bd k
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                    | k < k1 && k < bd  = goUp bd k1
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                    | otherwise        = k
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                      where
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                        k1 = fun k
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                goDown bd k
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                    | k1 < k && bd < k  = goDown bd k1
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                    | otherwise        = k
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                      where
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                        k1 = fun k
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                m0 = (l1 + h1) `div` 2
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                m1 = fun m0
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                lgs = takeWhile good [lo .. hi]
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                hgs = takeWhile good [hi,hi-1 .. lo]
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                mgs = reverse (takeWhile good [m0,m0-1 .. l1])
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                        ++ takeWhile good [m0+1 .. h1]
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problem_156=sum[sum $fixedPoints a|a<-[1..9]]
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</haskell>
+
 
+
== [http://projecteuler.net/index.php?section=problems&id=157 Problem 157] ==
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Solving the diophantine equation 1/a+1/b= p/10n
+
 
+
Solution:
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<haskell>
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problem_157 = undefined
+
</haskell>
+
 
+
== [http://projecteuler.net/index.php?section=problems&id=158 Problem 158] ==
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Exploring strings for which only one character comes lexicographically after its neighbour to the left.
+
 
+
Solution:
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<haskell>
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problem_158 = undefined
+
</haskell>
+
 
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== [http://projecteuler.net/index.php?section=problems&id=159 Problem 159] ==
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Digital root sums of factorisations.
+
 
+
Solution:
+
<haskell>
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import Control.Monad
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import Data.Array.ST
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import qualified Data.Array.Unboxed as U
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spfArray :: U.UArray Int Int
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spfArray  = runSTUArray (do
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  arr <- newArray (0,m-1) 0
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  loop arr 2
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  forM_ [2 .. m - 1] $ \ x ->
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    loop2 arr x 2
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  return arr
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  )
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  where
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  m=10^6
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  loop arr n
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      |n>=m=return ()
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      |otherwise=do writeArray arr n (n-9*(div (n-1) 9))
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                    loop arr (n+1)
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  loop2 arr x n
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      |n*x>=m=return ()
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      |otherwise=do incArray arr x n
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                    loop2 arr x (n+1)
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  incArray arr x n = do
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      a <- readArray arr x
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      b <- readArray arr n
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      ab <- readArray arr (x*n)
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      when(ab<a+b) (writeArray arr (x*n) (a + b))
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writ x=appendFile "p159.log"$foldl (++) "" [show x,"\n"]
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main=do
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    mapM_ writ $U.elems spfArray
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problem_159 = main
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--at first ,make main to get file "p159.log"
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--then ,add all num in the file
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</haskell>
+
 
+
== [http://projecteuler.net/index.php?section=problems&id=160 Problem 160] ==
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Factorial trailing digits
+
 
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We use the following two facts:
+
 
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Fact 1: <hask>(2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0</hask>
+
 
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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
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<hask>d == 5</hask>, and we can verify that directly by
+
evaluating the above two Haskell expressions.
+
 
+
More generally:
+
 
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Fact 1 follows from the fact that the group of invertible elements
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of the ring of integers modulo <hask>5^d</hask> has
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<hask>4*5^(d-1)</hask> elements.
+
 
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Fact 2 follows from the fact that the group of invertible elements
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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
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    multiply = foldl' times 1
+
    x `toPower` k = multiply $ genericReplicate n x
+
    e = facFactors 2 n - facFactors 5 n
+
    twos
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    | e <= d    = 2 `toPower` e
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    | otherwise = 2 `toPower` (d + (e - d) `mod` (4 * 5 ^ (d - 1)))
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    odds = multiply [odd | a <- takeWhile (<= n) $ iterate (* 2) 1,
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                          b <- takeWhile (<= n) $ iterate (* 5) a,
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                          odd <- [3, 5 .. n `div` b `mod` base],
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                          odd `mod` 5 /= 0]
+
 
+
-- The number of factors of the prime p in n!
+
facFactors p = sum . zipWith (*) (iterate (\x -> p * x + 1) 1) .
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              tail . radix p
+
 
+
-- The digits of n in base b representation
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radix p = map snd . takeWhile (/= (0, 0)) .
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          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
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mulMod a b c= (b * c) `rem` a
+
squareMod :: Integral a => a -> a -> a
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squareMod a b = (b * b) `rem` a
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pow' :: (Num a, Integral b) => (a -> a -> a) -> (a -> a) -> a -> b -> a
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pow' _ _ _ 0 = 1
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pow' mul sq x' n' = f x' n' 1
+
    where
+
    f x n y
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        | n == 1 = x `mul` y
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        | r == 0 = f x2 q y
+
        | otherwise = f x2 q (x `mul` y)
+
        where
+
            (q,r) = quotRem n 2
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            x2 = sq x
+
powMod :: Integral a => a -> a -> a -> a
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powMod m = pow' (mulMod m) (squareMod m)
+
+
productMod =foldl (mulMod (10^5)) 1
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hFacial 0=1
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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!