結果

問題 No.3018 目隠し宝探し
ユーザー kemuniku
提出日時 2025-01-25 13:39:03
言語 Nim
(2.2.0)
結果
AC  
実行時間 85 ms / 2,000 ms
コード長 4,416 bytes
コンパイル時間 5,583 ms
コンパイル使用メモリ 75,808 KB
実行使用メモリ 25,972 KB
平均クエリ数 2.59
最終ジャッジ日時 2025-01-25 22:53:47
合計ジャッジ時間 7,877 ms
ジャッジサーバーID
(参考情報)
judge4 / judge9
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 21
権限があれば一括ダウンロードができます

ソースコード

diff #

import macros;macro ImportExpand(s:untyped):untyped = parseStmt($s[2])
ImportExpand "cplib/tmpl/sheep.nim" <=== "when not declared CPLIB_TMPL_SHEEP:\n    const CPLIB_TMPL_SHEEP* = 1\n    {.warning[UnusedImport]: off.}\n    {.hint[XDeclaredButNotUsed]: off.}\n    import algorithm\n    import sequtils\n    import tables\n    import macros\n    import math\n    import sets\n    import strutils\n    import strformat\n    import sugar\n    import heapqueue\n    import streams\n    import deques\n    import bitops\n    import std/lenientops\n    import options\n    #入力系\n    proc scanf(formatstr: cstring){.header: \"<stdio.h>\", varargs.}\n    proc getchar(): char {.importc: \"getchar_unlocked\", header: \"<stdio.h>\", discardable.}\n    proc ii(): int {.inline.} = scanf(\"%lld\\n\", addr result)\n    proc lii(N: int): seq[int] {.inline.} = newSeqWith(N, ii())\n    proc si(): string {.inline.} =\n        result = \"\"\n        var c: char\n        while true:\n            c = getchar()\n            if c == ' ' or c == '\\n' or c == '\\255':\n                break\n            result &= c\n    #chmin,chmax\n    template `max=`(x, y) = x = max(x, y)\n    template `min=`(x, y) = x = min(x, y)\n    #bit演算\n    proc `%`*(x: int, y: int): int =\n        result = x mod y\n        if y > 0 and result < 0: result += y\n        if y < 0 and result > 0: result += y\n    proc `//`*(x: int, y: int): int{.inline.} =\n        result = x div y\n        if y > 0 and result * y > x: result -= 1\n        if y < 0 and result * y < x: result -= 1\n    proc `%=`(x: var int, y: int): void = x = x%y\n    proc `//=`(x: var int, y: int): void = x = x//y\n    proc `**`(x: int, y: int): int = x^y\n    proc `**=`(x: var int, y: int): void = x = x^y\n    proc `^`(x: int, y: int): int = x xor y\n    proc `|`(x: int, y: int): int = x or y\n    proc `&`(x: int, y: int): int = x and y\n    proc `>>`(x: int, y: int): int = x shr y\n    proc `<<`(x: int, y: int): int = x shl y\n    proc `~`(x: int): int = not x\n    proc `^=`(x: var int, y: int): void = x = x ^ y\n    proc `&=`(x: var int, y: int): void = x = x & y\n    proc `|=`(x: var int, y: int): void = x = x | y\n    proc `>>=`(x: var int, y: int): void = x = x >> y\n    proc `<<=`(x: var int, y: int): void = x = x << y\n    proc `[]`(x: int, n: int): bool = (x and (1 shl n)) != 0\n    #便利な変換\n    proc `!`(x: char, a = '0'): int = int(x)-int(a)\n    #定数\n    #[ include cplib/utils/constants ]#\n    when not declared CPLIB_UTILS_CONSTANTS:\n        const CPLIB_UTILS_CONSTANTS* = 1\n        const INF32*: int32 = 100100111.int32\n        const INF64*: int = int(3300300300300300491)\n    const INF = INF64\n    #converter\n\n    #range\n    iterator range(start: int, ends: int, step: int): int =\n        var i = start\n        if step < 0:\n            while i > ends:\n                yield i\n                i += step\n        elif step > 0:\n            while i < ends:\n                yield i\n                i += step\n    iterator range(ends: int): int = (for i in 0..<ends: yield i)\n    iterator range(start: int, ends: int): int = (for i in\n            start..<ends: yield i)\n\n    #joinが非stringでめちゃくちゃ遅いやつのパッチ\n    proc join*[T: not string](a: openArray[T], sep: string = \"\"): string = a.mapit($it).join(sep)\n\n    proc dump[T](arr:seq[seq[T]])=\n        for i in 0..<len(arr):\n            echo arr[i]\n"

# 3回聞くと,当然特定が可能

# ところで、それよりも小さい回数で特定が可能である例がある。
# たとえば、H = 1のとき、(1,1)からの距離を聞くことで1回の質問で宝物の場所を特定することができる。

# 右上と左上から聞けば、一意に特定できる。

var tmp = stdin.readline().split(" ").mapit(parseInt(it))
var H = tmp[0]
var W = tmp[1]
if H == 1 and W == 1:
    echo "! 1 1"
    quit()

echo "? ",1," ",1
var d1 = stdin.readLine().parseInt()

var kouho : seq[(int,int)]

for i in 1..H:
    for j in 1..W:
        if (i-1)**2 + (j-1)**2 == d1:
            kouho.add((i,j))

if len(kouho) == 1:
    echo "! ",kouho[0][0]," ",kouho[0][1]
    quit()

echo "? ",1," ",W
var d2 = stdin.readLine().parseInt()

var kouho2 : seq[(int,int)]
for (i,j) in kouho:
    if (i-1)**2 + (j-W)**2 == d2:
        kouho2.add((i,j))

if len(kouho2) == 1:
    echo "! ",kouho2[0][0]," ",kouho2[0][1]
    quit()
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