結果

問題 No.278 連続する整数の和(2)
ユーザー 👑 obakyanobakyan
提出日時 2019-06-02 22:50:18
言語 Lua
(LuaJit 2.1.1734355927)
結果
AC  
実行時間 41 ms / 2,000 ms
コード長 1,898 bytes
コンパイル時間 110 ms
コンパイル使用メモリ 6,816 KB
実行使用メモリ 20,036 KB
最終ジャッジ日時 2024-09-17 20:15:52
合計ジャッジ時間 1,030 ms
ジャッジサーバーID
(参考情報)
judge2 / judge4
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ファイルパターン 結果
other AC * 18
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ソースコード

diff #
プレゼンテーションモードにする

local mce, mfl, msq, mmi, mma = math.ceil, math.floor, math.sqrt, math.min, math.max
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
local asdf = {}
asdf.p = dv
asdf.cnt = 1
x = x / dv
while(x % dv == 0) do
x = x / dv
asdf.cnt = asdf.cnt + 1
end
table.insert(tmp, asdf)
lim = mce(msq(x))
end
if(primepos == prime_num) then break end
primepos = primepos + 1
dv = primes[primepos]
end
if(x ~= 1) then
local asdf = {}
asdf.p, asdf.cnt = x, 1
table.insert(tmp, asdf)
end
return tmp
end
local function getdivisor(divisorparts)
local t = {}
local pat = 1
local len = #divisorparts
local allpat = 1
for i = 1, len do
allpat = allpat * (1 + divisorparts[i].cnt)
end
for t_i_pat = 0, allpat - 1 do
local div = allpat
local i_pat = t_i_pat
local ret = 1
for i = 1, len do
div = mfl(div / (divisorparts[i].cnt + 1))
local mul = mfl(i_pat / div)
i_pat = i_pat % div
for j = 1, mul do
ret = ret * divisorparts[i].p
end
end
table.insert(t, ret)
end
-- table.sort(t)
return t
end
local n = io.read("*n")
if n % 2 == 0 then
n = mfl(n / 2)
end
local primes = getprimes(mce(msq(n)))
local divps = getdivisorparts(n, primes)
local divs = getdivisor(divps)
local ret = 0
for i = 1, #divs do
ret = ret + divs[i]
end
print(ret)
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0