#680012 (Lua) No.1597 Matrix Sort

• 問題: No.1597 Matrix Sort
• ユーザー: 👑 obakyan
• 提出日時: 2021-07-21 20:28:48
• 言語: Lua (LuaJit 2.1.1696795921)
• 結果: AC
• 実行時間: 290 ms / 1,500 ms
• コード長: 976 bytes
• コンパイル時間: 162 ms
• コンパイル使用メモリ: 5,120 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-07-17 14:17:10
• 合計ジャッジ時間: 6,784 ms
• ジャッジサーバーID （参考情報）: judge2 / judge1

入力	結果	実行時間; 実行使用メモリ
testcase_00	AC	1 ms; 5,248 KB
testcase_01	AC	2 ms; 5,376 KB
testcase_02	AC	2 ms; 5,376 KB
testcase_03	AC	290 ms; 5,376 KB
testcase_04	AC	280 ms; 5,376 KB
testcase_05	AC	287 ms; 5,376 KB
testcase_06	AC	276 ms; 5,376 KB
testcase_07	AC	283 ms; 5,376 KB
testcase_08	AC	219 ms; 5,376 KB
testcase_09	AC	221 ms; 5,376 KB
testcase_10	AC	124 ms; 5,376 KB
testcase_11	AC	135 ms; 5,376 KB
testcase_12	AC	160 ms; 5,376 KB
testcase_13	AC	235 ms; 5,376 KB
testcase_14	AC	248 ms; 5,376 KB
testcase_15	AC	137 ms; 5,376 KB
testcase_16	AC	153 ms; 5,376 KB
testcase_17	AC	165 ms; 5,376 KB
testcase_18	AC	275 ms; 5,376 KB
testcase_19	AC	179 ms; 5,376 KB
testcase_20	AC	155 ms; 5,376 KB
testcase_21	AC	147 ms; 5,376 KB
testcase_22	AC	144 ms; 5,376 KB
testcase_23	AC	135 ms; 5,376 KB
testcase_24	AC	108 ms; 5,376 KB
testcase_25	AC	105 ms; 5,376 KB
testcase_26	AC	2 ms; 5,376 KB
testcase_27	AC	1 ms; 5,376 KB
testcase_28	AC	1 ms; 5,376 KB
testcase_29	AC	2 ms; 5,376 KB

ソースコード

local mfl, mce = math.floor, math.ceil
local n, k, p = io.read("*n", "*n", "*n")
local a, b = {}, {}
for i = 1, n do
  a[i] = io.read("*n")
end
for i = 1, n do
  b[i] = io.read("*n")
end

table.sort(a)
table.sort(b)

local function solve(x)
  local c1 = 0
  local apos = n
  for i = 1, n do
    while 0 < apos and x < a[apos] + b[i] do
      apos = apos - 1
    end
    if apos == 0 then break end
    c1 = c1 + apos
  end
  apos = n
  local c2 = 0
  for i = 1, n do
    while 0 < apos and p - 1 < a[apos] + b[i] do
      apos = apos - 1
    end
    if apos == 0 then break end
    c2 = c2 + apos
  end
  apos = n
  local c3 = 0
  for i = 1, n do
    while 0 < apos and p + x < a[apos] + b[i] do
      apos = apos - 1
    end
    if apos == 0 then break end
    c3 = c3 + apos
  end
  return k <= c3 - c2 + c1
end
local min, max = -1, p - 1
while 1 < max - min do
  local mid = mfl((min + max) / 2)
  if solve(mid) then
    max = mid
  else
    min = mid
  end
end
print(max)