結果
| 問題 | No.1611 Minimum Multiple with Double Divisors |
| ユーザー |
 hitonanode
|
| 提出日時 | 2021-07-21 23:10:24 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 9,795 bytes |
| コンパイル時間 | 1,679 ms |
| コンパイル使用メモリ | 154,800 KB |
| 最終ジャッジ日時 | 2025-01-23 05:03:07 |
|
ジャッジサーバーID （参考情報） |
judge5 / judge3 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 27 WA * 10 |
ソースコード
#include <algorithm>#include <array>#include <bitset>#include <cassert>#include <chrono>#include <cmath>#include <complex>#include <deque>#include <forward_list>#include <fstream>#include <functional>#include <iomanip>#include <ios>#include <iostream>#include <limits>#include <list>#include <map>#include <numeric>#include <queue>#include <random>#include <set>#include <sstream>#include <stack>#include <string>#include <tuple>#include <type_traits>#include <unordered_map>#include <unordered_set>#include <utility>#include <vector>using namespace std;using lint = long long;using pint = pair<int, int>;using plint = pair<lint, lint>;struct fast_ios { fast_ios(){ cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); }; } fast_ios_;#define ALL(x) (x).begin(), (x).end()#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)#define REP(i, n) FOR(i,0,n)#define IREP(i, n) IFOR(i,0,n)template <typename T, typename V>void ndarray(vector<T>& vec, const V& val, int len) { vec.assign(len, val); }template <typename T, typename V, typename... Args> void ndarray(vector<T>& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); }template <typename T> bool chmax(T &m, const T q) { return m < q ? (m = q, true) : false; }template <typename T> bool chmin(T &m, const T q) { return m > q ? (m = q, true) : false; }int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); }template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }template <typename T> vector<T> sort_unique(vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end());return vec; }template <typename T> int arglb(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::lower_bound(v.begin(), v.end(), x)); }template <typename T> int argub(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::upper_bound(v.begin(), v.end(), x)); }template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }template <typename T, size_t sz> ostream &operator<<(ostream &os, const array<T, sz> &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']';return os; }#if __cplusplus >= 201703Ltemplate <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); returnis; }template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { os << '('; std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os << ')'; }#endiftemplate <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os;}template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }template <typename T, typename TH> ostream &operator<<(ostream &os, const unordered_set<T, TH> &vec) { os << '{'; for (auto v : vec) os << v << ',';os << '}'; return os; }template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os;}template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}';return os; }template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << '(' << pa.first << ',' << pa.second << ')';return os; }template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }template <typename TK, typename TV, typename TH> ostream &operator<<(ostream &os, const unordered_map<TK, TV, TH> &mp) { os << '{'; for (auto v : mp)os << v.first << "=>" << v.second << ','; os << '}'; return os; }#ifdef HITONANODE_LOCALconst string COLOR_RESET = "\033[0m", BRIGHT_GREEN = "\033[1;32m", BRIGHT_RED = "\033[1;31m", BRIGHT_CYAN = "\033[1;36m", NORMAL_CROSSED = "\033[0;9;37m", RED_BACKGROUND = "\033[1;41m", NORMAL_FAINT = "\033[0;2m";#define dbg(x) cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET <<endl#define dbgif(cond, x) ((cond) ? cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ <<COLOR_RESET << endl : cerr)#else#define dbg(x) (x)#define dbgif(cond, x) 0#endif// Linear sieve algorithm for fast prime factorization// Complexity: O(N) time, O(N) space:// - MAXN = 10^7: ~44 MB, 80~100 ms (Codeforces / AtCoder GCC, C++17)// - MAXN = 10^8: ~435 MB, 810~980 ms (Codeforces / AtCoder GCC, C++17)// Reference:// [1] D. Gries, J. Misra, "A Linear Sieve Algorithm for Finding Prime Numbers,"// Communications of the ACM, 21(12), 999-1003, 1978.// - https://cp-algorithms.com/algebra/prime-sieve-linear.html// - https://37zigen.com/linear-sieve/struct Sieve {std::vector<int> min_factor;std::vector<int> primes;Sieve(int MAXN) : min_factor(MAXN + 1) {for (int d = 2; d <= MAXN; d++) {if (!min_factor[d]) {min_factor[d] = d;primes.emplace_back(d);}for (const auto &p : primes) {if (p > min_factor[d] or d * p > MAXN) break;min_factor[d * p] = p;}}}// Prime factorization for 1 <= x <= MAXN^2// Complexity: O(log x) (x <= MAXN)// O(MAXN / log MAXN) (MAXN < x <= MAXN^2)template <typename T> std::map<T, int> factorize(T x) {std::map<T, int> ret;assert(x > 0 and x <= ((long long)min_factor.size() - 1) * ((long long)min_factor.size() - 1));for (const auto &p : primes) {if (x < T(min_factor.size())) break;while (!(x % p)) x /= p, ret[p]++;}if (x >= T(min_factor.size())) ret[x]++, x = 1;while (x > 1) ret[min_factor[x]]++, x /= min_factor[x];return ret;}// Enumerate divisors of 1 <= x <= MAXN^2// Be careful of highly composite numbers https://oeis.org/A002182/list https://gist.github.com/dario2994/fb4713f252ca86c1254d#file-list-txt// (n, (# of div. of n)): 45360->100, 735134400(<1e9)->1344, 963761198400(<1e12)->6720template <typename T> std::vector<T> divisors(T x) {std::vector<T> ret{1};for (const auto p : factorize(x)) {int n = ret.size();for (int i = 0; i < n; i++) {for (T a = 1, d = 1; d <= p.second; d++) {a *= p.first;ret.push_back(ret[i] * a);}}}return ret; // NOT sorted}// Moebius function Table, (-1)^{# of different prime factors} for square-free x// return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...] https://oeis.org/A008683std::vector<int> GenerateMoebiusFunctionTable() {std::vector<int> ret(min_factor.size());for (unsigned i = 1; i < min_factor.size(); i++) {if (i == 1)ret[i] = 1;else if ((i / min_factor[i]) % min_factor[i] == 0)ret[i] = 0;elseret[i] = -ret[i / min_factor[i]];}return ret;}// Calculate [0^K, 1^K, ..., nmax^K] in O(nmax)// Note: **0^0 == 1**template <typename MODINT> std::vector<MODINT> enumerate_kth_pows(long long K, int nmax) {assert(nmax < int(min_factor.size()));assert(K >= 0);if (K == 0) return std::vector<MODINT>(nmax + 1, 1);std::vector<MODINT> ret(nmax + 1);ret[0] = 0, ret[1] = 1;for (int n = 2; n <= nmax; n++) {if (min_factor[n] == n) {ret[n] = MODINT(n).pow(K);} else {ret[n] = ret[n / min_factor[n]] * ret[min_factor[n]];}}return ret;}};Sieve sieve(1 << 15); // (can factorize n <= 10^9)int cnt(lint x, lint p) {int ret = 0;while (x % p == 0) ret++, x /= p;return ret;}vector<map<int, int>> facs;lint solve(lint X) {lint ret = 1LL << 60;FOR(b, 2, facs.size()) {lint num = 1, den = 1;for (auto [p, deg] : facs[b]) {auto g = cnt(X, p);num *= deg + g + 1;den *= g + 1;lint gc = __gcd(num, den);num /= gc, den /= gc;if (num > den * 2) {num = den = 0;break;}}if (num == den * 2) chmin(ret, X * b);}return ret;}lint guchoku(lint X) {int n = sieve.divisors(X).size();for (lint Y = X; ; Y += X) {if (sieve.divisors(Y).size() == n * 2) return Y;}}int main() {facs.resize(60);FOR(i, 2, facs.size()) facs[i] = sieve.factorize(i);int T;cin >> T;while (T--) {lint X;cin >> X;cout << solve(X) << '\n';}}