Euler problems/151 to 160

Problem 151

Paper sheets of standard sizes: an expected-value problem.

Solution:

problem_151 = fun (1,1,1,1)
 
fun (0,0,0,1) = 0
fun (0,0,1,0) = fun (0,0,0,1) + 1
fun (0,1,0,0) = fun (0,0,1,1) + 1
fun (1,0,0,0) = fun (0,1,1,1) + 1
fun (a,b,c,d) = 
    (pickA + pickB + pickC + pickD) / (a + b + c + d)
    where
    pickA | a > 0 = a * fun (a-1,b+1,c+1,d+1)
          | otherwise = 0
    pickB | b > 0 = b * fun (a,b-1,c+1,d+1)
          | otherwise = 0
    pickC | c > 0 = c * fun (a,b,c-1,d+1)
          | otherwise = 0
    pickD | d > 0 = d * fun (a,b,c,d-1)
          | otherwise = 0

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]
 
 
-- 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 fun [(0, [[]])] [13, 7, 5]
    where
    fun 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 = 
    fun x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y)
    where
    fun 0 _        = [[]]
    fun x ((n, r, s):ns)
        | r > x     = fun x ns
        | s < x     = []
        | otherwise = map (n :) (fun (x - r) ns) ++ fun x ns
    fun _ _        = []
 
-- 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 = 
    [sort $ u ++ v |
    (x, s) <- only23,
    u <- findInvSq (1%2 - x) all23,
    v <- s
    ]
 
problem_152 = length solutions

Problem 153

Investigating Gaussian Integers

Solution:

#include <stdio.h>
#include <math.h>
typedef long long lolo;
static const lolo sumTo( lolo n ) { return n * ( n + 1 ) / 2; }

#define LL (1000)
lolo ssTab[ LL ];
int gcd(int a, int b) {
    if (b==0) return a;
    return gcd(b, a%b);
}

static const lolo sumSigma( lolo n ) {
    lolo a, r, s;

    if( n == 0 ) return 0;
    if( n < LL ) { r = ssTab[ n ]; if( r ) return r; }
    s = floor(sqrt( n ));
    r = 0;
    for( a = 1; a <= s; ++a ) r += a * ( n / a );
    for( a = 1; a <= s; ++a ) r += ( sumTo( n / a ) - sumTo ( n / ( a + 1 ) ) ) * a;
    if( n / s == s ) r -= s * s;
    if( n < LL ) ssTab[ n ] = r;

    return r;
}

int main() {
    const lolo m = 100000000;
    lolo t;
    int a, b;
    long ab;
    t = sumSigma(m);
    for( a = 1; a <=floor(sqrt(m)); ++a ) {
        for( b = 1; b <= a && a * a + b * b <= m; ++b ) {
            ab=(a*a+b*b);
            if( ( a | b ) & 1 && gcd( a, b ) == 1 ) {
                t += 2 * sumSigma( m / ab) * ( a == b ? a : a + b );
            }
        }
    }
    printf( "t = %lld\n", t );

    return 1;
}
problem_153 = main

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

Problem 155

Counting Capacitor Circuits.

Solution:

--http://www.research.att.com/~njas/sequences/A051389
a051389=
   [1, 2, 4, 8, 20, 42,
    102, 250, 610, 1486, 
    3710, 9228, 23050, 57718,
    145288, 365820, 922194, 2327914
   ]
problem_155 = sum a051389

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]]

Problem 157

Solving the diophantine equation 1/a+1/b= p/10n

Solution:

-- Call (a,b,p) a primitive tuple of equation 1/a+1/b=p/10^n
-- a and b are divisors of 10^n, gcd a b == 1, a <= b and a*b <= 10^n
-- I noticed that the number of variants with a primitive tuple
-- is equal to the number of divisors of p.
-- So I produced all possible primitive tuples per 10^n and
-- summed all the number of divisors of every p

import Data.List
k `deelt` n = n `mod` k == 0

delers n 
    | n == 10 = [1,2,5,10]
    | otherwise = 
        [ d |
        d <- [1..n `div` 5],
        d `deelt` n ] 
        ++ [n `div` 4, n `div` 2,n]
fp n = 
    [ n*(a+b) `div` ab |
    a <- ds,
    b <- dropWhile (<a) ds,
    gcd a b == 1,
    let ab = a*b,
    ab <= n 
    ]
    where
    ds = delers n
numDivisors :: Integer -> Integer
numDivisors n = product [ toInteger (a+1) | (p,a) <- primePowerFactors n]
numVgln = sum . map numDivisors . fp

main = do
    print . sum . map numVgln . takeWhile (<=10^9) . iterate (10*) $ 10 
primePowerFactors x = [(head a ,length a)|a<-group$primeFactors x] 
merge xs@(x:xt) ys@(y:yt) = case compare x y of
    LT -> x : (merge xt ys)
    EQ -> x : (merge xt yt)
    GT -> y : (merge xs yt)
    
diff  xs@(x:xt) ys@(y:yt) = case compare x y of
    LT -> x : (diff xt ys)
    EQ -> diff xt yt
    GT -> diff xs yt
 
primes, nonprimes :: [Integer]
primes    = [2,3,5] ++ (diff [7,9..] nonprimes) 
nonprimes = foldr1 f . map g $ tail primes
    where f (x:xt) ys = x : (merge xt ys)
          g p = [ n*p | n <- [p,p+2..]]
primeFactors n =
    factor n primes
    where
    factor n (p:ps) 
        | p*p > n        = [n]
        | n `mod` p == 0 = p : factor (n `div` p) (p:ps)
        | otherwise      = factor n ps

Problem 158

Exploring strings for which only one character comes lexicographically after its neighbour to the left.

Solution:

factorial n = product [1..toInteger n]
fallingFactorial x n = product [x - fromInteger i | i <- [0..toInteger n - 1] ]
choose n k = fallingFactorial n k `div` factorial k
fun n=(2 ^ n - n - 1) * choose 26 n 
problem_158=maximum$map fun [1..26]

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

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

Fact 2: 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, 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 5^d has 4*5^(d-1) elements.

Fact 2 follows from the fact that the group of invertible elements of the ring of integers modulo 10^d is isomorphic to the product 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)