結果
問題 | No.278 連続する整数の和(2) |
ユーザー |
👑 |
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 |
ソースコード
local mce, mfl, msq, mmi, mma = math.ceil, math.floor, math.sqrt, math.min, math.maxlocal function getprimes(x)local primes = {}local allnums = {}for i = 1, x do allnums[i] = true endfor i = 2, x doif(allnums[i]) thentable.insert(primes, i)local lim = mfl(x / i)for j = 2, lim doallnums[j * i] = falseendendendreturn primesendlocal function getdivisorparts(x, primes)local prime_num = #primeslocal tmp = {}local lim = mce(msq(x))local primepos = 1local dv = primes[primepos]while(primepos <= prime_num and dv <= lim) doif(x % dv == 0) thenlocal asdf = {}asdf.p = dvasdf.cnt = 1x = x / dvwhile(x % dv == 0) dox = x / dvasdf.cnt = asdf.cnt + 1endtable.insert(tmp, asdf)lim = mce(msq(x))endif(primepos == prime_num) then break endprimepos = primepos + 1dv = primes[primepos]endif(x ~= 1) thenlocal asdf = {}asdf.p, asdf.cnt = x, 1table.insert(tmp, asdf)endreturn tmpendlocal function getdivisor(divisorparts)local t = {}local pat = 1local len = #divisorpartslocal allpat = 1for i = 1, len doallpat = allpat * (1 + divisorparts[i].cnt)endfor t_i_pat = 0, allpat - 1 dolocal div = allpatlocal i_pat = t_i_patlocal ret = 1for i = 1, len dodiv = mfl(div / (divisorparts[i].cnt + 1))local mul = mfl(i_pat / div)i_pat = i_pat % divfor j = 1, mul doret = ret * divisorparts[i].pendendtable.insert(t, ret)end-- table.sort(t)return tendlocal n = io.read("*n")if n % 2 == 0 thenn = mfl(n / 2)endlocal primes = getprimes(mce(msq(n)))local divps = getdivisorparts(n, primes)local divs = getdivisor(divps)local ret = 0for i = 1, #divs doret = ret + divs[i]endprint(ret)