結果
問題 | No.1339 循環小数 |
ユーザー |
![]() |
提出日時 | 2021-01-15 21:44:14 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 699 ms / 2,000 ms |
コード長 | 6,676 bytes |
コンパイル時間 | 2,093 ms |
コンパイル使用メモリ | 200,456 KB |
最終ジャッジ日時 | 2025-01-17 18:55:01 |
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 36 |
ソースコード
#include <bits/stdc++.h>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) { if (m < q) {m = q; return true;} else return false; }template <typename T> bool chmin(T &m, const T q) { if (m > q) {m = q; return true;} else return 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> 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) { 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#else#define dbg(x) (x)#endif// Calculate log_A B (MOD M) (baby-step gian-step)// DiscreteLogarithm dl(M, A);// lint ans = dl.log(B);// Complexity: O(M^(1/2)) for each query// Verified: <https://judge.yosupo.jp/problem/discrete_logarithm_mod>// Constraints: 0 <= A < M, B < M, 1 <= M <= 1e9 (M is not limited to prime)struct DiscreteLogarithm {using lint = long long int;int M, stepsize;lint baby_a, giant_a, g;std::unordered_map<lint, int> baby_log_dict;lint inverse(lint a) {lint b = M / g, u = 1, v = 0;while (b) {lint t = a / b;a -= t * b;std::swap(a, b);u -= t * v;std::swap(u, v);}u %= M / g;return u >= 0 ? u : u + M / g;}DiscreteLogarithm(int mod, int a_new) : M(mod), baby_a(a_new % mod), giant_a(1) {g = 1;while (std::__gcd(baby_a, M / g) > 1) g *= std::__gcd(baby_a, M / g);stepsize = 32; // lg(MAX_M)while (stepsize * stepsize < M / g) stepsize++;lint now = 1 % (M / g), inv_g = inverse(baby_a);for (int n = 0; n < stepsize; n++) {if (!baby_log_dict.count(now)) baby_log_dict[now] = n;(now *= baby_a) %= M / g;(giant_a *= inv_g) %= M / g;}}// log(): returns the smallest nonnegative x that satisfies a^x = b mod M, or -1 if there's no solutionlint log(lint b) {b %= M;lint acc = 1 % M;for (int i = 0; i < stepsize; i++) {if (acc == b) return i;(acc *= baby_a) %= M;}if (b % g) return -1; // No solutionlint now = b * giant_a % (M / g);for (lint q = 1; q <= M / stepsize + 1; q++) {if (baby_log_dict.count(now)) return q * stepsize + baby_log_dict[now];(now *= giant_a) %= M / g;}return -1;}};// Solve ax+by=gcd(a, b)lint extgcd(lint a, lint b, lint &x, lint &y){lint d = a;if (b != 0) d = extgcd(b, a % b, y, x), y -= (a / b) * x;else x = 1, y = 0;return d;}// Calc a^(-1) (MOD m)lint mod_inverse(lint a, lint m){lint x, y;extgcd(a, m, x, y);return (m + x % m) % m;}lint solve(int N) {while (N % 2 == 0) N /= 2;while (N % 5 == 0) N /= 5;DiscreteLogarithm dl(N, 10);return dl.log(mod_inverse(10, N)) + 1;}int main() {int T;cin >> T;while (T--) {int N;cin >> N;cout << solve(N) << '\n';}}