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Euler problems/151 to 160

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Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_151 = undefined
+
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
 
</haskell>
 
</haskell>
   
Line 18: Line 18:
 
import Data.Ratio
 
import Data.Ratio
 
import Data.List
 
import Data.List
+
import Data.Ord (comparing)
  +
import Data.Function (on)
  +
 
invSq n = 1 % (n * n)
 
invSq n = 1 % (n * n)
 
sumInvSq = sum . map invSq
 
sumInvSq = sum . map invSq
+
 
subsets (x:xs) = let s = subsets xs in s ++ map (x :) s
 
subsets (x:xs) = let s = subsets xs in s ++ map (x :) s
 
subsets _ = [[]]
 
subsets _ = [[]]
+
 
primes = 2 : 3 : 7 : [p | p <- [11, 13..79],
 
primes = 2 : 3 : 7 : [p | p <- [11, 13..79],
 
all (\q -> p `mod` q /= 0) [3, 5, 7]]
 
all (\q -> p `mod` q /= 0) [3, 5, 7]]
+
 
-- All subsets whose sum of inverse squares,
 
-- All subsets whose sum of inverse squares,
 
-- when added to x, does not contain a factor of p
 
-- 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,
 
pfree s x p = [(y, t) | t <- subsets s, let y = x + sumInvSq t,
 
denominator y `mod` p /= 0]
 
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
 
-- All pairs (x, s) where x is a rational number whose reduced
 
-- denominator is not divisible by any prime greater than 3;
 
-- denominator is not divisible by any prime greater than 3;
 
-- and s is all sets of numbers up to 80 divisible
 
-- and s is all sets of numbers up to 80 divisible
 
-- by a prime greater than 3, whose sum of inverse squares is x.
 
-- by a prime greater than 3, whose sum of inverse squares is x.
only23 = foldl f [(0, [[]])] [13, 7, 5]
+
only23 = foldl fun [(0, [[]])] [13, 7, 5]
where
+
where
f a p = collect $ [(y, u ++ v) | (x, s) <- a,
+
fun a p =
(y, v) <- pfree (terms p) x p,
+
collect $ [(y, u ++ v) |
u <- s]
+
(x, s) <- a,
terms p = [n * p | n <- [1..80`div`p],
+
(y, v) <- pfree (terms p) x p,
all (\q -> n `mod` q /= 0) $
+
u <- s]
11 : takeWhile (>= p) [13, 7, 5]
+
terms p =
]
+
[n * p |
collect = map (\z -> (fst $ head z, map snd z)) .
+
n <- [1..80`div`p],
groupBy fstEq . sortBy cmpFst
+
all (\q -> n `mod` q /= 0) $
fstEq (x, _) (y, _) = x == y
+
11 : takeWhile (>= p) [13, 7, 5]
cmpFst (x, _) (y, _) = compare x y
+
]
+
collect =
  +
map (\z -> (fst $ head z, map snd z)) .
  +
groupBy fstEq . sortBy cmpFst
  +
fstEq = (==) `on` fst
  +
cmpFst = comparing fst
  +
 
-- All subsets (of an ordered set) whose sum of inverse squares is x
 
-- 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)
+
findInvSq x y =
where
+
fun x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y)
f 0 _ = [[]]
+
where
f x ((n, r, s):ns)
+
fun 0 _ = [[]]
| r > x = f x ns
+
fun x ((n, r, s):ns)
| s < x = []
+
| r > x = fun x ns
| otherwise = map (n :) (f (x - r) ns) ++ f x ns
+
| s < x = []
f _ _ = []
+
| otherwise = map (n :) (fun (x - r) ns) ++ fun x ns
+
fun _ _ = []
  +
 
-- All numbers up to 80 that are divisible only by the primes
 
-- All numbers up to 80 that are divisible only by the primes
 
-- 2 and 3 and are not divisible by 32 or 27.
 
-- 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]
 
all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 80]
+
solutions = if verify
+
solutions =
then [sort $ u ++ v | (x, s) <- only23,
+
[sort $ u ++ v |
u <- findInvSq (1%2 - x) all23,
+
(x, s) <- only23,
v <- s]
+
u <- findInvSq (1%2 - x) all23,
else undefined
+
v <- s
+
]
  +
 
problem_152 = length solutions
 
problem_152 = length solutions
 
</haskell>
 
</haskell>
Line 77: Line 77:
 
== [http://projecteuler.net/index.php?section=problems&id=153 Problem 153] ==
 
== [http://projecteuler.net/index.php?section=problems&id=153 Problem 153] ==
 
Investigating Gaussian Integers
 
Investigating Gaussian Integers
 
Solution:
 
<haskell>
 
problem_153 = undefined
 
</haskell>
 
   
 
== [http://projecteuler.net/index.php?section=problems&id=154 Problem 154] ==
 
== [http://projecteuler.net/index.php?section=problems&id=154 Problem 154] ==
 
Exploring Pascal's pyramid.
 
Exploring Pascal's pyramid.
   
Solution:
+
{{sect-stub}}
<haskell>
 
#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
 
</haskell>
 
   
 
== [http://projecteuler.net/index.php?section=problems&id=155 Problem 155] ==
 
== [http://projecteuler.net/index.php?section=problems&id=155 Problem 155] ==
Line 93: Line 88:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_155 = undefined
+
--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
 
</haskell>
 
</haskell>
   
 
== [http://projecteuler.net/index.php?section=problems&id=156 Problem 156] ==
 
== [http://projecteuler.net/index.php?section=problems&id=156 Problem 156] ==
 
Counting Digits
 
Counting Digits
 
Solution:
 
<haskell>
 
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]]
 
</haskell>
 
   
 
== [http://projecteuler.net/index.php?section=problems&id=157 Problem 157] ==
 
== [http://projecteuler.net/index.php?section=problems&id=157 Problem 157] ==
Line 172: Line 99:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_157 = undefined
+
-- 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 `divides` n = n `mod` k == 0
  +
  +
divisors n
  +
| n == 10 = [1,2,5,10]
  +
| otherwise =
  +
[ d |
  +
d <- [1..n `div` 5],
  +
d `divides` 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 = divisors 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
 
</haskell>
 
</haskell>
   
Line 180: Line 107:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_158 = undefined
+
factorial n = product [1..toInteger n]
  +
fallingFactorial x n = product [x - i | i <- [0..fromIntegral 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]
 
</haskell>
 
</haskell>
   
Line 193: Line 120:
 
spfArray :: U.UArray Int Int
 
spfArray :: U.UArray Int Int
 
spfArray = runSTUArray (do
 
spfArray = runSTUArray (do
arr <- newArray (0,m-1) 0
+
arr <- newArray (2,m-1) 0
loop arr 2
+
forM_ [2 .. m-1] $ \n ->
forM_ [2 .. m - 1] $ \ x ->
+
writeArray arr n (n-9*((n-1) `div` 9))
loop2 arr x 2
+
forM_ [2 .. m-1] $ \x ->
  +
forM_ [2 .. m`div`n-1] $ \n ->
  +
incArray arr x n
 
return arr
 
return arr
 
)
 
)
 
where
 
where
 
m=10^6
 
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
 
incArray arr x n = do
 
a <- readArray arr x
 
a <- readArray arr x
Line 214: Line 133:
 
ab <- readArray arr (x*n)
 
ab <- readArray arr (x*n)
 
when(ab<a+b) (writeArray arr (x*n) (a + b))
 
when(ab<a+b) (writeArray arr (x*n) (a + b))
writ x=appendFile "p159.log"$foldl (++) "" [show x,"\n"]
+
writ x=appendFile "p159.log"$ show x ++ "\n"
main=do
+
main=mapM_ writ $U.elems spfArray
mapM_ writ $U.elems spfArray
 
 
problem_159 = main
 
problem_159 = main
   
Line 298: Line 217:
 
hFacial 0=1
 
hFacial 0=1
 
hFacial a
 
hFacial a
|gcd a 5==1=mod (a*hFacial(a-1)) (5^5)
+
|gcd a 5==1=(a*hFacial(a-1)) `mod` (5^5)
 
|otherwise=hFacial(a-1)
 
|otherwise=hFacial(a-1)
fastFacial a= hFacial $mod a 6250
+
fastFacial a= hFacial $a `mod` 6250
numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]]
+
numPrime x p=takeWhile(>0) [x `div` (p^a)|a<-[1..]]
 
p160 x=mulMod t5 a b
 
p160 x=mulMod t5 a b
 
where
 
where
 
t5=10^5
 
t5=10^5
 
lst=numPrime x 5
 
lst=numPrime x 5
a=powMod t5 1563 $mod c 2500
+
a=powMod t5 1563 $c `mod` 2500
 
b=productMod c6
 
b=productMod c6
 
c=sum lst
 
c=sum lst

Latest revision as of 08:22, 23 February 2010

Contents

[edit] 1 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

[edit] 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
import Data.Ord (comparing)
import Data.Function (on)
 
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 = (==) `on` fst
    cmpFst = comparing fst
 
-- 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

[edit] 3 Problem 153

Investigating Gaussian Integers

[edit] 4 Problem 154

Exploring Pascal's pyramid.

[edit] 5 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

[edit] 6 Problem 156

Counting Digits

[edit] 7 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 `divides` n = n `mod` k == 0
 
divisors n 
    | n == 10 = [1,2,5,10]
    | otherwise = 
        [ d |
        d <- [1..n `div` 5],
        d `divides` 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 = divisors 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

[edit] 8 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 - i | i <- [0..fromIntegral 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]

[edit] 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 (2,m-1) 0 
  forM_ [2 .. m-1] $ \n ->
    writeArray arr n (n-9*((n-1) `div` 9))
  forM_ [2 .. m-1] $ \x ->
    forM_ [2 .. m`div`n-1] $ \n ->
      incArray arr x n
  return arr 
  )
  where
  m=10^6
  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"$ show x ++ "\n"
main=mapM_ writ $U.elems spfArray
problem_159 = main
 
--at first ,make main to get file "p159.log"
--then ,add all num in the file

[edit] 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
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=(a*hFacial(a-1)) `mod` (5^5)
    |otherwise=hFacial(a-1)
fastFacial a= hFacial $a `mod` 6250
numPrime x p=takeWhile(>0) [x `div` (p^a)|a<-[1..]]
p160 x=mulMod t5 a b
    where
    t5=10^5
    lst=numPrime x 5
    a=powMod t5 1563 $c `mod` 2500
    b=productMod  c6 
    c=sum lst
    c6=map fastFacial $x:lst
problem_160 = p160 (10^12)