結果
問題 | No.2724 Coprime Game 1 |
ユーザー |
|
提出日時 | 2024-09-25 03:05:42 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 3,911 bytes |
コンパイル時間 | 3,368 ms |
コンパイル使用メモリ | 256,560 KB |
実行使用メモリ | 157,752 KB |
最終ジャッジ日時 | 2024-09-25 03:05:52 |
合計ジャッジ時間 | 10,243 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | TLE * 1 |
other | -- * 7 |
ソースコード
#include <bits/stdc++.h>using namespace std;#define rep(i, n) for(int i=0; i<n; i++)#define debug 0#define YES cout << "Yes" << endl;#define NO cout << "No" << endl;using ll = long long;using ld = long double;const int mod = 998244353;const int MOD = 1000000007;const double pi = atan2(0, -1);const int inf = 1 << 31-1;const ll INF = 1LL << 63 - 1;#include <time.h>#include <chrono>//vectorの中身を空白区切りで出力template<typename T>void printv(vector<T> v) {for (int i = 0; i < v.size(); i++) {cout << v[i];if (i < v.size() - 1) {cout << " ";}}cout << endl;}//vectorの中身を改行区切りで出力template<typename T>void print1(vector<T> v) {for (auto x : v) {cout << x << endl;}}//二次元配列を出力template<typename T>void printvv(vector<vector<T>> vv) {for (vector<T> v : vv) {printv(v);}}//vectorを降順にソートtemplate<typename T>void rsort(vector<T>& v) {sort(v.begin(), v.end());reverse(v.begin(), v.end());}//昇順priority_queueを召喚template<typename T>struct rpriority_queue {priority_queue<T, vector<T>, greater<T>> pq;void push(T x) {pq.push(x);}void pop() {pq.pop();}T top() {return pq.top();}size_t size() {return pq.size();}bool empty() {return pq.empty();}};//mod mod下で逆元を算出する//高速a^n計算(mod ver.)ll power(ll a, ll n) {if (n == 0) {return 1;}else if (n % 2 == 0) {ll x = power(a, n / 2);x *= x;x %= mod;return x;}else {ll x = power(a, n - 1);x *= a;x %= mod;return x;}}//フェルマーの小定理を利用ll modinv(ll p) {return power(p, mod - 2) % mod;}//Mexを求めるstruct Mex {map<int, int> mp;set<int> s;Mex(int Max) {for (int i = 0; i <= Max; i++) {s.insert(i);}}int _mex = 0;void Input(int x) {mp[x]++;s.erase(x);if (_mex == x) {_mex = *begin(s);}}void Remove(int x) {if (mp[x] == 0) {cout << "Mex ERROR!: NO VALUE WILL BE REMOVED" << endl;}mp[x]--;if (mp[x] == 0) {s.insert(x);if (*begin(s) == x) {_mex = x;}}}int mex(){return _mex;}};//Union-Findstruct UnionFind {vector<int> par;UnionFind(int N) : par(N) {for (int i = 0; i < N; i++) {par[i] = -1;}}//root(x):xの根を求める関数int root(int x) {if (par[x] == -1) {return x;}else {return par[x] = root(par[x]);}}//isSame(x,y):xとyが同じグループならtrueを返す関数bool isSame(int x, int y) {if (root(x) == root(y)) {return true;}else {return false;}}//Union(x,y):xとyの根をつなげる関数void Union(int x, int y) {int X = root(x);int Y = root(y);if (X == Y) {return;}if (X < Y) {swap(X, Y);}par[X] = Y;}};int main() {int T;cin >> T;vector<int> prime;set<int> list;set<int> s;for (int i = 2; i <= 3000000; i++) {s.insert(i);}while (!s.empty()) {int a = *begin(s);//cout << "new prime=" << a << endl;prime.push_back(a);list.insert(a);s.erase(a);for (ll k = a; k * (ll)a <= 3000000; k++) {s.erase(k * a);}}prime.push_back(mod);int M = prime.size();rep(t, T) {int N;cin >> N;if (list.count(N)) {cout << "P" << endl;}else {int b = N/2;int u = N;int lb = 0, rb = M;int lu = 0, ru = M;while (lb + 1 < rb) {int mid = (lb + rb) / 2;if (prime[mid] <= b) {lb = mid;}else {rb = mid;}}while (lu + 1 < ru) {int mid = (lu + ru) / 2;if (prime[mid] <= u) {lu = mid;}else {ru = mid;}}//cout << "prime["<<lb<<"]=" << prime[lb] << " prime["<<lu<<"]=" << prime[lu] << endl;int p_cnt = lu - lb;if ((N - p_cnt) % 2 == 0) {cout << "P" << endl;}else {cout << "K" << endl;}}}}