## SPOJ 6560. N DIV PHI_N (Hard)

SPOJ 6560. N DIV PHI_N (Hard) also accepted using the same idea accept using a bit memoization. Precalculation of product of primes in prime table 10^25000 and linear search is enough to get accepted.Its first 6000 primes whose products are sufficient . Accepted Haskell code. I also tried a slightly improved Rabin Miller primality testing with idea give topcoder forum for faster modular exponentiation for large numbers and code with isProbable function got accepted with 3.95 and time limit exceed other time.

import Data.List import Data.Maybe import Data.Bits import qualified Data.ByteString.Char8 as BS powM::Integer->Integer->Integer->Integer powM a d n | d == 0 = 1 | d == 1 = mod a n | otherwise = mod q n where p = powM ( mod ( a^2 ) n ) ( shiftR d 1 ) n q = if (.&.) d 1 == 1 then mod ( a * p ) n else p powL :: Integer -> Integer -> Integer -> Integer powL a d n | d <= 10^10 = powM a d n | otherwise = mod ( powM ( powL a ( div d 1000000 ) n ) 1000000 n * powM a ( mod d 1000000 ) n ) n isProbable::Integer-> Bool isProbable n |n<=1 = False |n==2 = True |even n = False |otherwise = rabinMiller [2,3] n s d where s = until ( \x -> testBit ( n - 1) ( fromIntegral x ) ) ( +1 ) 0 d = div ( n - 1 ) ( shiftL 1 ( fromIntegral s ) ) --d=until (\x->mod x 2/=0) (`div` 2) (n-1) --(n-1=2^s*d) --s=until (\x->d*2^x==pred n) (+1) 0 --counts the power rabinMiller::[Integer]->Integer->Integer->Integer-> Bool rabinMiller [] _ _ _ = True rabinMiller (a:xs) n s d = if a == n then True else case x==1 of True -> rabinMiller xs n s d _ -> if any ( == pred n) . take ( fromIntegral s ) . iterate ( \e -> mod (e^2) n ) $ x then rabinMiller xs n s d else False where x=powL a d n prime :: [Integer] prime = 2:3:filter isPrime [2*k+1 | k<-[2..]] isPrime :: Integer -> Bool isPrime n = all ((/=0). mod n ).takeWhile ( <= ( truncate.sqrt.fromIntegral $ n) ) $ prime anstable ::[Integer] anstable = scanl1 (*) prime helpSolve :: [Integer] -> Integer -> Integer -> Integer helpSolve (x:xs) n ret | x > n = ret | otherwise = helpSolve xs n x solve :: Integer -> BS.ByteString solve 1 = BS.pack "1" solve n = BS.pack.show $ a where a = helpSolve ( tail anstable ) n ( head anstable ) readInt :: BS.ByteString -> Integer readInt = fst.fromJust.BS.readInteger main = BS.interact $ BS.unlines.map ( solve .readInt ) . tail . BS.lines

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## 2 Comments »

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Looks a little too heavy in my opinion. Why not just use sieve of Eratosthenes from haskellwiki? Although anyway nice.

Comment by gorlum0 | July 26, 2011 |

Yes its bit heavy because of Rabin Miller code. Thanks for the suggestion for sieve of Eratosthenes. It did not occur to me to solve this problem using array.

Comment by tiwari_mukesh | July 26, 2011 |