#649766 (C++14) No.1113 二つの整数 / Two Integers

• 問題: No.1113 二つの整数 / Two Integers
• ユーザー: karinohito
• 提出日時: 2021-04-19 22:19:40
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 2 ms / 1,000 ms
• コード長: 1,815 bytes
• コンパイル時間: 960 ms
• コンパイル使用メモリ: 104,904 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-07-04 05:09:35
• 合計ジャッジ時間: 1,764 ms
• ジャッジサーバーID （参考情報）: judge4 / judge1

ファイルパターン	結果
sample	AC * 2
other	AC * 15

ソースコード

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//#include <atcoder/all>
#include <iostream>
#include <numeric>
#include <cmath>
#include <limits>
#include <stdio.h>
#include <iomanip>
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
using namespace std;
//using namespace atcoder;
using ll = long long;
#define all(A) A.begin(),A.end()
using vll = vector<ll>;
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
using Graph = vector<vector<ll>>;
vector<ll> fact, factinv, inv;
ll mod = 1e9 + 7;
void prenCkModp(ll n) {
    fact.resize(n + 5);
    factinv.resize(n + 5);
    inv.resize(n + 5);
    fact.at(0) = fact.at(1) = 1;
    factinv.at(0) = factinv.at(1) = 1;
    inv.at(1) = 1;
    for (ll i = 2; i < n + 5; i++) {
        fact.at(i) = (fact.at(i - 1) * i) % mod;
        inv.at(i) = mod - (inv.at(mod % i) * (mod / i)) % mod;
        factinv.at(i) = (factinv.at(i - 1) * inv.at(i)) % mod;
    }
}
ll nCk(ll n, ll k) {
    if (n < k) return 0;
    return fact.at(n) * (factinv.at(k) * factinv.at(n - k) % mod) % mod;
}
ll gcd(ll(a), ll(b)) {
    ll c = a;
    while (a % b != 0) {
        c = a % b;
        a = b;
        b = c;
    }
    return b;
}
int main() {
    ll X, Y;
    cin >> X >> Y;
    ll G = gcd(X, Y);
    ll D = sqrt(G);
    bool C = true;
    for (ll d = D - 1000; d < D + 1000; d++) {
        if (d * d == G)C = false;
    }
    cout << (C ? "Even" : "Odd") << endl;
}
 
 
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