結果

問題 No.1113 二つの整数 / Two Integers
ユーザー kyon2326
提出日時 2020-07-18 11:37:33
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 26 ms / 1,000 ms
コード長 9,430 bytes
コンパイル時間 12,106 ms
コンパイル使用メモリ 438,708 KB
最終ジャッジ日時 2025-01-12 00:15:11
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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 * 2
other AC * 15
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

#include <bits/stdc++.h>
using namespace std;
#if __has_include(<boost/multiprecision/cpp_int.hpp>)
    #include <boost/multiprecision/cpp_int.hpp>
    #include <boost/multiprecision/cpp_dec_float.hpp>
    using bll = boost::multiprecision::cpp_int;
    using bdouble = boost::multiprecision::number<boost::multiprecision::cpp_dec_float<100>>;
    using namespace boost::multiprecision;
#endif
#if __has_include(<boost/stacktrace.hpp>)
    #define BOOST_STACKTRACE_USE_ADDR2LINE
    #define BOOST_STACKTRACE_ADDR2LINE_LOCATION /usr/local/opt/binutils/bin/addr2line
    #define _GNU_SOURCE 1
    #include <boost/stacktrace.hpp>
#endif
#ifdef LOCAL_TEST
    namespace std {
        template<typename T> class dvector : public std::vector<T> {
        public:
            dvector() : std::vector<T>() {}
            explicit dvector(size_t n, const T& value = T()) : std::vector<T>(n, value) {}
            dvector(const std::vector<T>& v) : std::vector<T>(v) {}
            dvector(const std::initializer_list<T> il) : std::vector<T>(il) {}
            dvector(const std::string::iterator first, const std::string::iterator last) : std::vector<T>(first, last) {}
            dvector(const typename std::vector<T>::iterator first, const typename std::vector<T>::iterator last) : std::vector<T>(first, last) {}
            dvector(const typename std::vector<T>::reverse_iterator first, const typename std::vector<T>::reverse_iterator last) : std::vector<T
                >(first, last) {}
            dvector(const typename std::vector<T>::const_iterator first, const typename std::vector<T>::const_iterator last) : std::vector<T>(first, 
                last) {}
            dvector(const typename std::vector<T>::const_reverse_iterator first, const typename std::vector<T>::const_reverse_iterator last) : std
                ::vector<T>(first, last) {}
            T& operator[](size_t n) {
                try { return this->at(n); } catch (const std::exception& e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; return this->at(n
                    ); }
            }
            const T& operator[](size_t n) const {
                try { return this->at(n); } catch (const std::exception& e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; return this->at(n
                    ); }
            }
        };
    }
    class dbool {
    private:
        bool boolvalue;
    public:
        dbool() : boolvalue(false) {}
        dbool(bool b) : boolvalue(b) {}
        operator bool&() { return boolvalue; }
        operator const bool&() const { return boolvalue; }
    };
    #define vector dvector
    #define bool dbool
#endif
#ifdef LOCAL_DEV
    template<typename T> std::ostream& operator<<(std::ostream& s, const std::vector<T>& v) {
        for (size_t i = 0; i < v.size(); ++i){ s << v[i]; if (i < v.size() - 1) s << "\t"; } return s; }
    template<typename T> std::ostream& operator<<(std::ostream& s, const std::vector<std::vector<T>>& vv) {
        s << "\\\n"; for (size_t i = 0; i < vv.size(); ++i){ s << vv[i] << "\n"; } return s; }
    template<typename T> std::ostream& operator<<(std::ostream& s, const std::deque<T>& v) {
        for (size_t i = 0; i < v.size(); ++i){ s << v[i]; if (i < v.size() - 1) s << "\t"; } return s; }
    template<typename T> std::ostream& operator<<(std::ostream& s, const std::set<T>& se) {
        s << "{ "; for (auto itr = se.begin(); itr != se.end(); ++itr){ s << (*itr) << "\t"; } s << "}"; return s; }
    template<typename T> std::ostream& operator<<(std::ostream& s, const std::multiset<T>& se) {
        s << "{ "; for (auto itr = se.begin(); itr != se.end(); ++itr){ s << (*itr) << "\t"; } s << "}"; return s; }
    template <typename T, size_t N> std::ostream& operator<<(std::ostream& s, const std::array<T, N>& a) {
        s << "{ "; for (size_t i = 0; i < N; ++i){ s << a[i] << "\t"; } s << "}"; return s; }
    template<typename T1, typename T2> std::ostream& operator<<(std::ostream& s, const std::map<T1, T2>& m) {
        s << "{\n"; for (auto itr = m.begin(); itr != m.end(); ++itr){ s << "\t" << (*itr).first << " : " << (*itr).second << "\n"; } s << "}"; 
            return s; }
    template<typename T1, typename T2> std::ostream& operator<<(std::ostream& s, const std::pair<T1, T2>& p) {
        return s << "(" << p.first << ", " << p.second << ")"; }
    class SIGFPE_exception : std::exception {};
    class SIGSEGV_exception : std::exception {};
    void catch_SIGFPE([[maybe_unused]] int e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; throw SIGFPE_exception(); }
    void catch_SIGSEGV([[maybe_unused]] int e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; throw SIGSEGV_exception(); }
    signed convertedmain();
    signed main() { signal(SIGFPE, catch_SIGFPE); signal(SIGSEGV, catch_SIGSEGV); return convertedmain(); }
    #define main() convertedmain()
    void debug_impl() { std::cerr << '\n'; }
    template<typename Head, typename... Tail> void debug_impl(Head head, Tail... tail) { std::cerr << " " << head << (sizeof...(tail) ? "," : ""); 
        debug_impl(tail...); }
    #define debug(...) do { std::cerr << "(" << #__VA_ARGS__ << ") ="; debug_impl(__VA_ARGS__); } while (false)
#else
    #define debug(...) do {} while (false)
#endif
//#define int long long
using ll = long long;
//constexpr int INF = (ll)1e9 + 7;//INT_MAX=(1<<31)-1=2147483647
constexpr ll INF = (ll)1e18;//(1LL<<63)-1=9223372036854775807
constexpr ll MOD = (ll)1e9 + 7;
constexpr double EPS = 1e-9;
constexpr ll dx[4] = {1, 0, -1, 0};
constexpr ll dy[4] = {0, 1, 0, -1};
constexpr ll dx8[8] = {1, 0, -1, 0, 1, 1, -1, -1};
constexpr ll dy8[8] = {0, 1, 0, -1, 1, -1, 1, -1};
#define rep(i, n)   for(ll i=0, i##_length=(n); i< i##_length; ++i)
#define repeq(i, n) for(ll i=1, i##_length=(n); i<=i##_length; ++i)
#define rrep(i, n)   for(ll i=(n)-1; i>=0; --i)
#define rrepeq(i, n) for(ll i=(n)  ; i>=1; --i)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
void p() { std::cout << '\n'; }
template<typename Head, typename... Tail> void p(Head head, Tail... tail) { std::cout << head << (sizeof...(tail) ? " " : ""); p(tail...); }
template<typename T> inline void pv(std::vector<T>& v) { for(ll i=0, N=v.size(); i<N; i++) std::cout << v[i] << " \n"[i==N-1]; }
template<typename T> inline bool chmax(T& a, T b) { return a < b && (a = b, true); }
template<typename T> inline bool chmin(T& a, T b) { return a > b && (a = b, true); }
template<typename T> inline void uniq(std::vector<T>& v) { v.erase(std::unique(v.begin(), v.end()), v.end()); }
/*-----8<-----template-----8<-----*/
//ミラーラビン素数判定法とポラード・ロー法のコードは http://quiz.fuqinho.net/blog/2012/06/12/poj-2429-gcd-and-lcm-inverse/ より
// return (a * b) % m
ll mod_mult(ll a, ll b, ll m) {
    ll res = 0;
    ll exp = a % m;
    while (b) {
        if (b & 1) {
            res += exp;
            if (res > m) res -= m;
        }
        exp <<= 1;
        if (exp > m) exp -= m;
        b >>= 1;
    }
    return res;
}
// return (a ^ b) % m
ll mod_exp(ll a, ll b, ll m) {
    ll res = 1;
    ll exp = a % m;
    while (b) {
        if (b & 1) res = mod_mult(res, exp, m);
        exp = mod_mult(exp, exp, m);
        b >>= 1;
    }
    return res;
}
// ミラー-ラビン素数判定法
bool miller_rabin(ll n, ll times) {
    if (n < 2) return false;
    if (n == 2) return true;
    if (!(n & 1)) return false;
 
    ll q = n - 1;
    int k = 0;
    while (q % 2 == 0) {
        k++;
        q >>= 1;
    }
    // n - 1 = 2^k * q (qは奇素数)
    // nが素数であれば、下記のいずれかを満たす
    // (i) a^q ≡ 1 (mod n)
    // (ii) a^q, a^2q,..., a^(k-1)q のどれかがnを法として-1
    //
    // なので、逆に(i)(ii)いずれも満たしていない時は合成数と判定できる
    //
    for (int i = 0; i < times; i++) {
        ll a = rand() % (n - 1) + 1; // 1,..,n-1からランダムに値を選ぶ
        ll x = mod_exp(a, q, n);
        // (i)をチェック
        if (x == 1) continue;
        // (ii)をチェック
        bool found = false;
        for (int j = 0; j < k; j++) {
            if (x == n - 1) {
                found = true;
                break;
            }
            x = mod_mult(x, x, n);
        }
        if (found) continue;
 
        return false;
    }
    return true;
}
ll get_gcd(ll n, ll m) {
    if (n < m) swap(n, m);
    while (m) {
        swap(n, m);
        m %= n;
    }
    return n;
}
// ポラード・ロー素因数分解法
ll pollard_rho(ll n, int c) {
    ll x = 2;
    ll y = 2;
    ll d = 1;
    while (d == 1) {
        x = mod_mult(x, x, n) + c;
        y = mod_mult(y, y, n) + c;
        y = mod_mult(y, y, n) + c;
        d = get_gcd((x - y >= 0 ? x - y : y - x), n);
    }
    if (d == n) return pollard_rho(n, c + 1);
    return d;
}
vector<int> primes;
vector<bool> is_prime;
// 小さい素数(MAX_PRIMEまで)は先に用意しとく
void init_primes(ll MAX_PRIME) {
    is_prime.assign(MAX_PRIME + 1, true);
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i <= MAX_PRIME; i++) {
        if (is_prime[i]) {
            primes.push_back(i);
            for (int j = i * 2; j <= MAX_PRIME; j += i) {
                is_prime[j] = false;
            }
        }
    }
}
 
// 素数かどうか判定。大きければミラーラビンを使う
bool isPrime(ll n) {
    if (n < (ll)is_prime.size()) return is_prime[n];
    else return miller_rabin(n, 7);
}
 
// 素因数分解する。小さい数は用意した素数で試し割り、大きければポラード・ロー
void factorize(ll n, map<ll, ll>& factors) {
    if (isPrime(n)) {
        factors[n]++;
    }
    else {
        for (int i = 0; i < (int)primes.size(); i++) {
            int p = primes[i];
            while (n % p == 0) {
                factors[p]++;
                n /= p;
            }
        }
        if (n != 1) {
            if (isPrime(n)) {
                factors[n]++;
            }
            else {
                ll d = pollard_rho(n, 1);
                factorize(d, factors);
                factorize(n / d, factors);
            }
        }
    }
}
/*-----8<-----library-----8<-----*/
void solve() {
    init_primes(100);
    ll A,B;
    cin>>A>>B;
    ll g=gcd(A,B);
    map<ll,ll> m;
    factorize(g,m);
    for(auto&& x:m){
        if((x.second+1)%2==0){
            p("Even");return;
        }
    }
    p("Odd");
}
signed main() {
    solve();
    return 0;
}
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