結果

問題 No.153 石の山
ユーザー ecotteaecottea
提出日時 2021-08-24 19:27:27
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 5,000 ms
コード長 7,129 bytes
コンパイル時間 4,136 ms
コンパイル使用メモリ 230,024 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-11-15 07:34:22
合計ジャッジ時間 5,459 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 27
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#ifndef HIDDEN_IN_VISUAL_STUDIO //
//
#define _CRT_SECURE_NO_WARNINGS
// 使
#include <bits/stdc++.h>
#include <functional> // function
#include <fstream> // ifstream
using namespace std;
//
using ll = long long; // -2^63 2^63 = 9 * 10^18int -2^31 2^31 = 2 * 10^9
using pii = pair<int, int>; using pll = pair<ll, ll>; using pil = pair<int, ll>; using pli = pair<ll, int>;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using vl = vector<ll>; using vvl = vector<vl>; using vvvl = vector<vvl>;
using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;
using vc = vector<char>; using vvc = vector<vc>; using vvvc = vector<vvc>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
template <class T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
//
const double PI = 3.14159265359; //
const double DEG = PI / 180.0; // θ [deg] = θ * DEG [rad]
const vi dx4 = { 1, 0, -1, 0 }; // 4
const vi dy4 = { 0, 1, 0, -1 };
const vi dx8 = { 1, 1, 0, -1, -1, -1, 0, 1 }; // 8
const vi dy8 = { 0, 1, 1, 1, 0, -1, -1, -1 };
const ll INFL = (ll)3e18; const int INF = (int)1e9;
const double EPS = 1e-10; // 調
//
#define all(a) (a).begin(), (a).end()
#define sz(x) ((int)(x).size())
#define Yes(b) {cout << ((b) ? "Yes" : "No") << endl;}
#define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 n-1
#define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s t
#define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s t
#define repe(v, a) for(const auto& v : (a)) // a
#define repea(v, a) for(auto& v : (a)) // a
#define repb(set, d) for(int set = 0; set < (1 << int(d)); ++set) // d
#define repbm(mid, set, d) for(int mid = set; mid < (1 << int(d)); mid = (mid + 1) | set) // set
#define repbs(sub, set) for (int sub = set, bsub = 1; bsub > 0; bsub = sub, sub = (sub - 1) & set) // set
#define repbc(set, k, d) for (int set = (1 << k) - 1, lb, nx; set < (1 << n); lb = set & -set, nx = set + lb, set = (((set & ~nx) / lb) >> 1) | nx)
    // k
#define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a
#define repit(it, a) for(auto it = (a).begin(); it != (a).end(); ++it) //
#define repitr(it, a) for(auto it = (a).rbegin(); it != (a).rend(); ++it) //
//
inline ll pow(ll n, int k) { ll v = 1; rep(i, k) v *= n; return v; }
inline ll pow(int n, int k) { ll v = 1; rep(i, k) v *= n; return v; }
template <class T> inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // true
    
template <class T> inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // true
    
// >>, <<
template <class T, class U> inline istream& operator>> (istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template <class T, class U> inline ostream& operator<< (ostream& os, const pair<T, U>& p) { os << "(" << p.first << "," << p.second << ")"; return os
    ; }
template <class T, class U, class V> inline istream& operator>> (istream& is, tuple<T, U, V>& t) { is >> get<0>(t) >> get<1>(t) >> get<2>(t); return
    is; }
template <class T, class U, class V> inline ostream& operator<< (ostream& os, const tuple<T, U, V>& t) { os << "(" << get<0>(t) << "," << get<1>(t)
    << "," << get<2>(t) << ")"; return os; }
template <class T, class U, class V, class W> inline istream& operator>> (istream& is, tuple<T, U, V, W>& t) { is >> get<0>(t) >> get<1>(t) >> get<2
    >(t) >> get<3>(t); return is; }
template <class T, class U, class V, class W> inline ostream& operator<< (ostream& os, const tuple<T, U, V, W>& t) { os << "(" << get<0>(t) << "," <<
    get<1>(t) << "," << get<2>(t) << "," << get<3>(t) << ")"; return os; }
template <class T> inline istream& operator>> (istream& is, vector<T>& v) { repea(x, v) is >> x; return is; }
template <class T> inline ostream& operator<< (ostream& os, const vector<T>& v) { repe(x, v) os << x << " "; return os; }
template <class T> inline ostream& operator<< (ostream& os, const set<T>& s) { repe(x, s) os << x << " "; return os; }
template <class T> inline ostream& operator<< (ostream& os, const unordered_set<T>& s) { repe(x, s) os << x << " "; return os; }
template <class T, class U> inline ostream& operator<< (ostream& os, const map<T, U>& m) { repe(p, m) os << p << " "; return os; }
// Visual Studio
#ifdef _MSC_VER
#define popcount (int)__popcnt // 1
#define popcountll (int)__popcnt64
inline int ctz(unsigned int n) { unsigned long i; _BitScanForward(&i, n); return i; } // 0
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } //
#define dump(x) cerr << "[DEBUG]" << endl << x << endl; //
#define dumpel(v) cerr << "[DEBUG]" << endl; repe(x, v) {cerr << x << endl;}
#define dumpeli(v) cerr << "[DEBUG]" << endl; rep (i, sz(v)) {cerr << i << ": " << v[i] << endl;}
// GCC
#else
#define popcount (int)__builtin_popcount
#define popcountll (int)__builtin_popcountll
#define ctz __builtin_ctz
#define gcd __gcd
#define dump(x)
#define dumpel(v)
#define dumpeli(v)
#endif
#endif //
//-----------------AtCoder -----------------
#include <atcoder/all>
using namespace atcoder;
using mint = modint1000000007;
//using mint = modint998244353;
//using mint = modint; // mint::set_mod(m);
istream& operator>> (istream& is, mint& x) { ll tmp; is >> tmp; x = tmp; return is; }
ostream& operator<< (ostream& os, const mint& x) { os << x.val(); return os; }
using vm = vector<mint>; using vvm = vector<vm>; using vvvm = vector<vvm>;
//----------------------------------------------
int main() {
// std::ifstream in("input.txt"); std::cin.rdbuf(in.rdbuf()); //
cout << fixed << setprecision(15); //
int n;
cin >> n;
vi dp(n + 1);
dp[1] = 0;
repi(i, 2, n) {
int v2 = 0, v3 = 0;
if (i >= 2) {
v2 = dp[i / 2] ^ dp[(i + 1) / 2];
}
if (i >= 3) {
v3 = dp[i / 3] ^ dp[(i + 1) / 3] ^ dp[(i + 2) / 3];
}
if (v2 != 0 && v3 != 0) {
dp[i] = 0;
}
else if (v2 != 1 && v3 != 1) {
dp[i] = 1;
}
else {
dp[i] = 2;
}
}
cout << (dp[n] ? "A" : "B") << endl;
}
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