結果
問題 |
No.1739 Princess vs. Dragoness (& AoE)
|
ユーザー |
👑 |
提出日時 | 2021-11-13 20:39:39 |
言語 | Lua (LuaJit 2.1.1734355927) |
結果 |
TLE
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 2,056 bytes |
コンパイル時間 | 437 ms |
コンパイル使用メモリ | 5,504 KB |
実行使用メモリ | 17,312 KB |
最終ジャッジ日時 | 2024-11-27 22:02:30 |
合計ジャッジ時間 | 63,522 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 32 TLE * 8 |
ソースコード
local mfl, mce = math.floor, math.ceil local mmi, mma = math.min, math.max local bls, brs = bit.lshift, bit.rshift local Heapq = {} Heapq.create = function(self, lt) self.lt = lt self.cnt = 0 self.t = {} end Heapq.push = function(self, v) local hqlt = self.lt local hqt = self.t local c = self.cnt + 1 self.cnt = c hqt[c] = v while 1 < c do local p = brs(c, 1) if hqlt(hqt[c], hqt[p]) then hqt[c], hqt[p] = hqt[p], hqt[c] c = p else break end end end Heapq.pop = function(self) local hqlt = self.lt local hqt = self.t local ret = hqt[1] local c = self.cnt hqt[1] = hqt[c] c = c - 1 self.cnt = c local p = 1 while true do local d1, d2 = p * 2, p * 2 + 1 if c < d1 then break elseif c < d2 then if hqlt(hqt[d1], hqt[p]) then hqt[d1], hqt[p] = hqt[p], hqt[d1] end break else if hqlt(hqt[d1], hqt[d2]) then if hqlt(hqt[d1], hqt[p]) then hqt[d1], hqt[p] = hqt[p], hqt[d1] p = d1 else break end else if hqlt(hqt[d2], hqt[p]) then hqt[d2], hqt[p] = hqt[p], hqt[d2] p = d2 else break end end end end return ret end Heapq.new = function(lt) local obj = {} setmetatable(obj, {__index = Heapq}) obj:create(lt) return obj end local n = io.read("*n") local a, b = io.read("*n", "*n") local x, y = io.read("*n", "*n") local hs = {} for i = 1, n do table.insert(hs, io.read("*n")) end local function solve(z) local hq = Heapq.new(function(c, d) return c > d end) for i = 1, n do local h = hs[i] hq:push(h) end for i = 1, a do local v = hq:pop() if v <= z then hq:push(v) break end v = mma(z, v - x) hq:push(v) end local tot = 0 for i = 1, n do tot = tot + mma(0, hq:pop() - z) end return tot <= b * y end local min, max = -1, 1000000007 while 1 < max - min do local mid = mfl((min + max) / 2) if solve(mid) then max = mid else min = mid end end print(max)