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Commit 8b3f7f7

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ABC284-D
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‎README.md‎

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@@ -386,3 +386,16 @@ Haskellで競技プログラミングをやるテクニックは「[Haskellで
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* [ ] B - メタ構文変数
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* [ ] C - データ構造
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* [x] D - 見たことのない多項式
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## AtCoder Beginner Contest 284
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<https://atcoder.jp/contests/abc284>
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* [ ] A - Sequence of Strings
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* [ ] B - Multi Test Cases
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* [ ] C - Count Connected Components
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* [x] D - Happy New Year 2023
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* [ ] E - Count Simple Paths
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* [ ] F - ABCBAC
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* [ ] G - Only Once
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* [ ] Ex - Count Unlabeled Graphs

‎abc/abc284-d/Main.hs‎

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-- https://github.com/minoki/my-atcoder-solutions
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{-# LANGUAGE TypeApplications #-}
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{-# LANGUAGE BangPatterns #-}
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import Data.Int (Int64)
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import Control.Monad
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import qualified Data.Vector.Unboxed as U
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import qualified Data.Vector.Unboxed.Mutable as UM
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{-
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-- i <= 4.5 * 10^18
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integerSquareRoot :: Int64 -> Int64
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integerSquareRoot i = loop 2 2121320343
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where loop !low !high | low == high = low
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| otherwise = let !mid = (low + high) `quot` 2
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!mid2 = mid * mid
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in case compare i mid2 of
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LT -> loop low mid
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EQ -> mid
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GT -> loop mid high
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-}
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integerSquareRoot :: Int64 -> Int64
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integerSquareRoot = round . sqrt . fromIntegral
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solve :: [Int64] -> Int64 -> (Int64, Int64)
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solve primes !n = case [(p,m) | p <- primes, (m,0) <- [n `quotRem` p]] of
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[] -> error "No prime factor found"
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(p,m):_ -> if m `rem` p == 0 then
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(p, m `quot` p)
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else
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-- m <= 4.5 * 10^18
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(integerSquareRoot m, p)
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main = do
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t <- readLn @Int
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tests <- U.replicateM t $ readLn @Int64
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let !primes = sieve 2080083 -- 2080083^3 < 9*10^18 < 2080084^3
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U.forM_ tests $ \t -> do
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let (p, q) = solve primes t
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putStrLn $ show p ++ " " ++ show q
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--
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-- Sieve of Eratosthenes
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--
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infixr 5 !:
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(!:) :: a -> [a] -> [a]
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(!x) !: xs = x : xs
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-- | エラトステネスの篩により、 max 以下の素数の一覧を構築して返す
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-- >>> sieve 100
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-- [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
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sieve :: Int -> [Int64]
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sieve !max = 2 : U.ifoldr (\i isPrime xs -> if isPrime then fromIntegral (2 * i + 1) !: xs else xs) [] vec
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where
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vec = U.create $ do
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vec <- UM.replicate ((max - 1) `quot` 2 + 1) True
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UM.write vec 0 False -- 1 is not a prime
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-- vec ! i : is (2 * i + 1) prime?
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let clear !p = forM_ [3*p,5*p..max] $ \n -> UM.write vec (n `quot` 2) False
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factorBound = floor (sqrt (fromIntegral max) :: Double)
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loop !i | 2 * i + 1 > factorBound = return ()
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| otherwise = do b <- UM.read vec i
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when b $ clear (2 * i + 1)
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loop (i + 1)
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loop 1
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return vec

‎abc/abc284-d/Search.hs‎

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import Data.Int (Int64)
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import qualified Test.QuickCheck as QC
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import Math.NumberTheory.Primes
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test :: Int64 -> Bool
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test n = truncate (sqrt (fromIntegral (n^2 :: Int64) :: Double)) < n
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examples :: [Int64]
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examples = [a | p <- [nextPrime 94906266..precPrime 3037000499], let a = fromInteger (unPrime p), test a]
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prop :: QC.Property
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prop = let gen = QC.choose (94906266, 3037000499) -- 94906265^2 < 2^53 < 94906266^2, 3037000499^2 < 2^63 - 1 < 3037000500^2
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in QC.forAll gen (\n -> truncate (fromIntegral (n^2 :: Int64) :: Double) < n^2)
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-- main = QC.quickCheck prop
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main = print $ take 10 [n | n <- [94906266..3037000499], test n]
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-- main = print examples

‎abc/abc284-d/main-binarysearch.c‎

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#include <inttypes.h>
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#include <math.h>
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#include <stdio.h>
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#include <stdlib.h>
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// 入力:n <= 4.5*10^18, nは平方数
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int64_t isqrt(int64_t n)
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{
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// 4.5*10^18 < 2^62
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// 真の答えは 2 <= _ < 2^31 の範囲にある
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int64_t low = 2, high = INT64_C(1) << 31;
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while (low < high) {
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int64_t mid = (low + high) / 2;
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int64_t mid2 = mid * mid;
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if (mid2 < n) {
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low = mid;
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} else if (mid2 == n) {
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return mid;
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} else {
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high = mid;
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}
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}
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return low;
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}
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struct result {
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int64_t p, q;
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};
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// 入力:N <= 9*10^18
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struct result solve(int64_t N)
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{
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// 2080083^3 < 9*10^18 < 2080084^3
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for (int64_t a = 2; a <= 2080083; ++a) {
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if (N % a == 0) {
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int64_t b = N / a;
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if (b % a == 0) {
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// a = p
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return (struct result){.p = a, .q = b / a};
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} else {
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// a = q
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return (struct result){.p = isqrt(b), .q = a};
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}
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}
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}
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abort();
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}
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int main()
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{
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int T;
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scanf("%d", &T);
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for (int i = 0; i < T; ++i) {
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int64_t N;
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scanf("%" SCNd64, &N);
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struct result r = solve(N);
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printf("%" PRId64 " %" PRId64 "\n", r.p, r.q);
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}
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}

‎abc/abc284-d/main-longdouble.c‎

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#include <assert.h>
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#include <float.h>
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#include <inttypes.h>
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#include <math.h>
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#include <stdio.h>
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#include <stdlib.h>
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static_assert(LDBL_MANT_DIG >= 64, "not enough precision");
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// 入力:n <= 4.5*10^18, nは平方数
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int64_t isqrt(int64_t n)
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{
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return (int64_t)sqrtl((long double)n);
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}
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struct result {
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int64_t p, q;
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};
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// 入力:N <= 9*10^18
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struct result solve(int64_t N)
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{
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// 2080083^3 < 9*10^18 < 2080084^3
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for (int64_t a = 2; a <= 2080083; ++a) {
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if (N % a == 0) {
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int64_t b = N / a;
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if (b % a == 0) {
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// a = p
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return (struct result){.p = a, .q = b / a};
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} else {
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// a = q
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return (struct result){.p = isqrt(b), .q = a};
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}
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}
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}
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abort();
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}
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int main()
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{
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int T;
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scanf("%d", &T);
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for (int i = 0; i < T; ++i) {
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int64_t N;
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scanf("%" SCNd64, &N);
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struct result r = solve(N);
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printf("%" PRId64 " %" PRId64 "\n", r.p, r.q);
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}
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}

‎abc/abc284-d/main-round.c‎

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#include <inttypes.h>
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#include <math.h>
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#include <stdio.h>
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#include <stdlib.h>
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// 入力:n <= 4.5*10^18, nは平方数
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int64_t isqrt(int64_t n)
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{
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return llround(sqrt((double)n));
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}
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struct result {
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int64_t p, q;
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};
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// 入力:N <= 9*10^18
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struct result solve(int64_t N)
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{
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// 2080083^3 < 9*10^18 < 2080084^3
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for (int64_t a = 2; a <= 2080083; ++a) {
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if (N % a == 0) {
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int64_t b = N / a;
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if (b % a == 0) {
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// a = p
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return (struct result){.p = a, .q = b / a};
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} else {
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// a = q
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return (struct result){.p = isqrt(b), .q = a};
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}
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}
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}
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abort();
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}
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int main()
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{
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int T;
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scanf("%d", &T);
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for (int i = 0; i < T; ++i) {
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int64_t N;
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scanf("%" SCNd64, &N);
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struct result r = solve(N);
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printf("%" PRId64 " %" PRId64 "\n", r.p, r.q);
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}
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}

‎abc/abc284-d/main-trunc.c‎

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#include <inttypes.h>
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#include <math.h>
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#include <stdio.h>
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#include <stdlib.h>
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// 入力:n <= 4.5*10^18, nは平方数
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int64_t isqrt(int64_t n)
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{
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return (int64_t)sqrt((double)n);
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}
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struct result {
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int64_t p, q;
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};
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// 入力:N <= 9*10^18
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struct result solve(int64_t N)
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{
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// 2080083^3 < 9*10^18 < 2080084^3
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for (int64_t a = 2; a <= 2080083; ++a) {
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if (N % a == 0) {
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int64_t b = N / a;
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if (b % a == 0) {
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// a = p
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return (struct result){.p = a, .q = b / a};
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} else {
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// a = q
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return (struct result){.p = isqrt(b), .q = a};
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}
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}
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}
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abort();
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}
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int main()
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{
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int T;
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scanf("%d", &T);
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for (int i = 0; i < T; ++i) {
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int64_t N;
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scanf("%" SCNd64, &N);
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struct result r = solve(N);
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printf("%" PRId64 " %" PRId64 "\n", r.p, r.q);
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}
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}

‎abc/abc284-d/validate-sqrt-trunc.c‎

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#include <inttypes.h>
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#include <math.h>
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#include <stdio.h>
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int main()
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{
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// 3037000499^2 < 2^63-1 < 3037000500^2
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for (int64_t i = 0; i <= 3037000499; ++i) {
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int64_t n = i * i;
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int64_t j = (int64_t)sqrt((double)n);
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if (i != j) {
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printf("%" PRId64 "\n", i);
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return 0;
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}
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}
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puts("Done");
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}

‎package.yaml‎

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- unboxing-vector
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- vector-algorithms
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- QuickCheck
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- arithmoi
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ghc-options:
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# Maximum heap size: 1GiB

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