Euler problems/151 to 160
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1 Problem 151
Paper sheets of standard sizes: an expected-value problem.
Solution:
problem_151 = undefined
2 Problem 152
Writing 1/2 as a sum of inverse squares
Note that if p is an odd prime, the sum of inverse squares of all terms divisible by p must have reduced denominator not divisible by p.
Solution:
import Data.Ratio import Data.List 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
3 Problem 153
Investigating Gaussian Integers
Solution:
problem_153 = undefined
4 Problem 154
Exploring Pascal's pyramid.
Solution:
#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
5 Problem 155
Counting Capacitor Circuits.
Solution:
problem_155 = undefined
6 Problem 156
Counting Digits
Solution:
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]]
7 Problem 157
Solving the diophantine equation 1/a+1/b= p/10n
Solution:
problem_157 = undefined
8 Problem 158
Exploring strings for which only one character comes lexicographically after its neighbour to the left.
Solution:
problem_158 = undefined
9 Problem 159
Digital root sums of factorisations.
Solution:
import Control.Monad import Data.Array.ST import qualified Data.Array.Unboxed as U spfArray :: U.UArray Int Int spfArray = runSTUArray (do arr <- newArray (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
10 Problem 160
Factorial trailing digits
We use the following two facts:
Fact 1:(2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0
product [n | n <- [0..10^d], gcd n 10 == 1] `mod` 10^d == 1
We really only need these two facts for the special case of
d == 5
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 modulo5^d
4*5^(d-1)
Fact 2 follows from the fact that the group of invertible elements
of the ring of integers modulo10^d
of a cyclic group of order 2 and another cyclic group.
Solution:
problem_160 = trailingFactorialDigits 5 (10^12) trailingFactorialDigits d n = twos `times` odds where base = 10 ^ d x `times` y = (x * y) `mod` base multiply = foldl' times 1 x `toPower` k = multiply $ genericReplicate n x e = facFactors 2 n - facFactors 5 n twos | e <= d = 2 `toPower` e | otherwise = 2 `toPower` (d + (e - d) `mod` (4 * 5 ^ (d - 1))) odds = multiply [odd | a <- takeWhile (<= n) $ iterate (* 2) 1, b <- takeWhile (<= n) $ iterate (* 5) a, odd <- [3, 5 .. n `div` b `mod` base], odd `mod` 5 /= 0] -- The number of factors of the prime p in n! facFactors p = sum . zipWith (*) (iterate (\x -> p * x + 1) 1) . tail . radix p -- The digits of n in base b representation radix p = map snd . takeWhile (/= (0, 0)) . iterate ((`divMod` p) . fst) . (`divMod` p)
it have another fast way to do this .
Solution:
import Data.List mulMod :: Integral a => a -> a -> a -> a mulMod a b c= (b * c) `rem` a squareMod :: Integral a => a -> a -> a squareMod a b = (b * b) `rem` a pow' :: (Num a, Integral b) => (a -> a -> a) -> (a -> a) -> a -> b -> a pow' _ _ _ 0 = 1 pow' mul sq x' n' = f x' n' 1 where f x n y | n == 1 = x `mul` y | r == 0 = f x2 q y | otherwise = f x2 q (x `mul` y) where (q,r) = quotRem n 2 x2 = sq x powMod :: Integral a => a -> a -> a -> a powMod m = pow' (mulMod m) (squareMod m) productMod =foldl (mulMod (10^5)) 1 hFacial 0=1 hFacial a |gcd a 5==1=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)
