Difference between revisions of "Euler problems/151 to 160"
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+ | Do them on your own! |
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− | == [http://projecteuler.net/index.php?section=problems&id=151 Problem 151] == |
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− | 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> |
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− | |||
− | == [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 |
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− | |||
− | Solution: |
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− | <haskell> |
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− | problem_157 = undefined |
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− | </haskell> |
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− | |||
− | == [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. |
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− | |||
− | Solution: |
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− | <haskell> |
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− | problem_158 = undefined |
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− | </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. |
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− | |||
− | Solution: |
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− | <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> |
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− | |||
− | == [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> |
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− | |||
− | 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 |
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− | evaluating the above two Haskell expressions. |
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− | |||
− | 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 |
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− | of a cyclic group of order 2 and another cyclic group. |
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− | |||
− | Solution: |
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− | <haskell> |
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− | problem_160 = trailingFactorialDigits 5 (10^12) |
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− | |||
− | trailingFactorialDigits d n = twos `times` odds |
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− | where |
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− | base = 10 ^ d |
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− | x `times` y = (x * y) `mod` base |
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− | multiply = foldl' times 1 |
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− | x `toPower` k = multiply $ genericReplicate n x |
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− | e = facFactors 2 n - facFactors 5 n |
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− | 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] |
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− | |||
− | -- The number of factors of the prime p in n! |
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− | facFactors p = sum . zipWith (*) (iterate (\x -> p * x + 1) 1) . |
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− | tail . radix p |
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− | |||
− | -- 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) |
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− | </haskell> |
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− | it have another fast way to do this . |
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− | |||
− | Solution: |
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− | <haskell> |
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− | import Data.List |
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− | mulMod :: Integral a => a -> a -> a -> a |
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− | mulMod a b c= (b * c) `rem` a |
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− | 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 |
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− | where |
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− | f x n y |
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− | | n == 1 = x `mul` y |
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− | | r == 0 = f x2 q y |
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− | | otherwise = f x2 q (x `mul` y) |
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− | where |
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− | (q,r) = quotRem n 2 |
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− | x2 = sq x |
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− | powMod :: Integral a => a -> a -> a -> a |
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− | powMod m = pow' (mulMod m) (squareMod m) |
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− | |||
− | productMod =foldl (mulMod (10^5)) 1 |
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− | hFacial 0=1 |
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− | hFacial a |
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− | |gcd a 5==1=mod (a*hFacial(a-1)) (5^5) |
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− | |otherwise=hFacial(a-1) |
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− | fastFacial a= hFacial $mod a 6250 |
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− | numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]] |
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− | p160 x=mulMod t5 a b |
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− | where |
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− | t5=10^5 |
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− | lst=numPrime x 5 |
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− | a=powMod t5 1563 $mod c 2500 |
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− | b=productMod c6 |
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− | c=sum lst |
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− | c6=map fastFacial $x:lst |
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− | problem_160 = p160 (10^12) |
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− | |||
− | </haskell> |
Revision as of 21:43, 29 January 2008
Do them on your own!