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!

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