結果
問題 |
No.1611 Minimum Multiple with Double Divisors
|
ユーザー |
👑 |
提出日時 | 2021-07-21 23:35:34 |
言語 | Lua (LuaJit 2.1.1734355927) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,131 bytes |
コンパイル時間 | 408 ms |
コンパイル使用メモリ | 6,816 KB |
実行使用メモリ | 21,056 KB |
最終ジャッジ日時 | 2024-07-17 20:42:59 |
合計ジャッジ時間 | 7,254 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | -- * 2 |
other | TLE * 1 -- * 36 |
ソースコード
local mce, mfl, msq, mmi, mma, mab = math.ceil, math.floor, math.sqrt, math.min, math.max, math.abs local function getprimes(x) local primes = {} local allnums = {} for i = 1, x do allnums[i] = true end for i = 2, x do if allnums[i] then table.insert(primes, i) local lim = mfl(x / i) for j = 2, lim do allnums[j * i] = false end end end return primes end local function getdivisorparts(x, primes) local prime_num = #primes local tmp = {} local lim = mce(msq(x)) local primepos = 1 local dv = primes[primepos] while primepos <= prime_num and dv <= lim do if x % dv == 0 then x = mfl(x / dv) local cnt = 1 while x % dv == 0 do x = mfl(x / dv) cnt = cnt + 1 end table.insert(tmp, {dv, cnt}) lim = mce(msq(x)) end if primepos == prime_num then break end primepos = primepos + 1 dv = primes[primepos] end if x ~= 1 then table.insert(tmp, {x, 1}) end return tmp end local primes = getprimes(400000) local q = io.read("*n") for iq = 1, q do local x = io.read("*n") local v = false for i = 1, #primes do local p = primes[i] if x % p ~= 0 then v = p break end end local ans = 1 * v ans = ans * x local dvp = getdivisorparts(x, primes) local tot = 1 for i = 1, #dvp do tot = tot * (1 + dvp[i][2]) end local box = {} if 1 < x then box[1] = dvp[#dvp][2] for i = 2, #dvp do box[i] = box[i - 1] * dvp[#dvp + 1 - i][2] end end local function DIG(i, rem, v) local p = dvp[i][1] local z = dvp[i][2] if i == #dvp then if z + 1 <= rem then for j = 1, rem - 1 do v = v * p end if v < ans then ans = v end end else for j = 1, z - 1 do v = v * p end for j = z, 2 * z + 1 do v = v * p local qqq = mfl(rem / (j + 1)) if qqq < box[#dvp - i] then break end if rem % (j + 1) == 0 then DIG(i + 1, qqq, v) end end end end if 1 < x then DIG(1, tot * 2, 1) end print(ans) end