結果
問題 | No.1140 EXPotentiaLLL! |
ユーザー | hitonanode |
提出日時 | 2020-08-08 21:54:20 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 6,285 bytes |
コンパイル時間 | 2,302 ms |
コンパイル使用メモリ | 212,588 KB |
実行使用メモリ | 25,132 KB |
最終ジャッジ日時 | 2024-10-02 08:03:01 |
合計ジャッジ時間 | 7,909 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 233 ms
24,872 KB |
testcase_01 | AC | 242 ms
24,876 KB |
testcase_02 | WA | - |
testcase_03 | AC | 233 ms
24,796 KB |
testcase_04 | AC | 196 ms
24,876 KB |
testcase_05 | AC | 266 ms
24,872 KB |
testcase_06 | WA | - |
testcase_07 | WA | - |
testcase_08 | AC | 97 ms
24,876 KB |
testcase_09 | WA | - |
testcase_10 | WA | - |
testcase_11 | WA | - |
testcase_12 | AC | 86 ms
24,744 KB |
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> void ndarray(vector<T> &vec, int len) { vec.resize(len); } template <typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); } template <typename V, typename T> void ndfill(V &x, const T &val) { x = val; } template <typename V, typename T> void ndfill(vector<V> &vec, const T &val) { for (auto &v : vec) ndfill(v, val); } 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; } 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> srtunq(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> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); 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> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os; } template <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> ostream &operator<<(ostream &os, const unordered_set<T> &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> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; } #ifdef HITONANODE_LOCAL #define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl #else #define dbg(x) #endif // Sieve of Eratosthenes // (*this)[i] = (divisor of i, greater than 1) // Example: [0, 1, 2, 3, 2, 5, 3, 7, 2, 3, 2, 11, ...] // Complexity: Space O(MAXN), Time (construction) O(MAXNloglogMAXN) struct SieveOfEratosthenes : std::vector<int> { std::vector<int> primes; SieveOfEratosthenes(int MAXN) : std::vector<int>(MAXN + 1) { std::iota(begin(), end(), 0); for (int i = 2; i <= MAXN; i++) { if ((*this)[i] == i) { primes.push_back(i); for (int j = i; j <= MAXN; j += i) (*this)[j] = i; } } } using T = long long int; // Prime factorization for x <= MAXN^2 // Complexity: O(log x) (x <= MAXN) // O(MAXN / logMAXN) (MAXN < x <= MAXN^2) std::map<T, int> Factorize(T x) { assert(x <= 1LL * (int(size()) - 1) * (int(size()) - 1)); std::map<T, int> ret; if (x < int(size())) { while (x > 1) { ret[(*this)[x]]++; x /= (*this)[x]; } } else { for (auto p : primes) { while (!(x % p)) x /= p, ret[p]++; if (x == 1) break; } if (x > 1) ret[x]++; } return ret; } std::vector<T> Divisors(T x) { std::vector<T> ret{1}; for (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 // return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...] std::vector<int> GenerateMoebiusFunctionTable() { std::vector<int> ret(size()); for (int i = 1; i < int(size()); i++) { if (i == 1) ret[i] = 1; else if ((i / (*this)[i]) % (*this)[i] == 0) ret[i] = 0; else ret[i] = -ret[i / (*this)[i]]; } return ret; } }; SieveOfEratosthenes sieve(5000000); lint power(lint x, lint n, lint MOD) { lint ans = 1; while (n>0) { if (n & 1) (ans *= x) %= MOD; (x *= x) %= MOD; n >>= 1; } return ans; } int main() { int T; cin >> T; while (T--) { lint A, P; cin >> A >> P; if (sieve[P] == P and P != 1) { cout << 1 << '\n'; } else { cout << -1 << '\n'; } } }