結果
問題 | No.811 約数の個数の最大化 |
ユーザー |
|
提出日時 | 2020-09-09 20:54:54 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 37 ms / 2,000 ms |
コード長 | 1,572 bytes |
コンパイル時間 | 1,670 ms |
コンパイル使用メモリ | 169,048 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-12-16 06:54:04 |
合計ジャッジ時間 | 2,575 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 12 |
コンパイルメッセージ
In file included from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/istream:39, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/sstream:38, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/complex:45, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/ccomplex:39, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/x86_64-pc-linux-gnu/bits/stdc++.h:54, from main.cpp:2: In member function 'std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long long int) [with _CharT = char; _Traits = std::char_traits<char>]', inlined from 'int main()' at main.cpp:71:13: /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/ostream:202:25: warning: 'ans' may be used uninitialized [-Wmaybe-uninitialized] 202 | { return _M_insert(__n); } | ~~~~~~~~~^~~~~ main.cpp: In function 'int main()': main.cpp:59:16: note: 'ans' was declared here 59 | ll en = 0, ans; | ^~~
ソースコード
#include <algorithm>#include <bits/stdc++.h>using namespace std;#define rep(i, n) for (int i = 0; i < (int)(n); i++)#define rrep(i, n) for (int i = (int)n - 1; i > -1; i--)#define all(x) (x).begin(), (x).end()#define ll long long#define ld long double#define INF 1000000000000000000typedef pair<ll, ll> pll;int N, K;pll prime_factorize(long long N) {ll res1 = 0, res2 = 1;for (long long a = 2; a * a <= N; ++a) {if (N % a != 0)continue;long long ex = 0; // 指数// 割れる限り割り続けるwhile (N % a == 0) {++ex;N /= a;}// その結果を pushres1 += ex;res2 *= (ex + 1);}// 最後に残った数についてif (N != 1) {res1 += 1;res2 *= 2;}return {res1, res2};}long long GCD(long long a, long long b) {if (b == 0)return a;elsereturn GCD(b, a % b);}int main() {cin.tie(0);ios::sync_with_stdio(false);cin >> N >> K;vector<ll> tmp;for (int i = 1; i < N; i++) {int gcd = GCD(i, N);pll pri = prime_factorize(gcd);if (pri.first >= K)tmp.push_back(i);}ll en = 0, ans;rep(i, tmp.size()) {pll pri = prime_factorize(tmp[i]);if (pri.second > en) {en = pri.second;ans = tmp[i];} else if (pri.second == en) {if (tmp[i] < ans)ans = tmp[i];}}cout << ans << endl;return 0;}