結果

問題 No.2790 Athena 3
ユーザー au7777
提出日時 2024-07-04 01:18:15
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,061 bytes
コンパイル時間 4,852 ms
コンパイル使用メモリ 245,252 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-04 01:18:21
合計ジャッジ時間 5,547 ms
ジャッジサーバーID
（参考情報） 		judge1 / 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 * 3
other AC * 14
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>
#include <atcoder/all>
typedef long long int ll;
using namespace std;
typedef pair<ll, ll> P;
using namespace atcoder;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
#define USE998244353
#ifdef USE998244353
const ll MOD = 998244353;
using mint = modint998244353;
#else
const ll MOD = 1000000007;
using mint = modint1000000007;
#endif
#pragma region //使いがち
const int MAX = 2000001;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll gcd(ll x, ll y) {
   if (y == 0) return x;
   else if (y > x) {
       return gcd (y, x); 
   }
   else return gcd(x % y, y);
}
ll lcm(ll x, ll y) {
   return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
    ll m = 0;
    ll M = 3000000001;
    while (M - m > 1) {
        ll now = (M + m) / 2;
        if (now * now <= x) {
            m = now;
        }
        else {
            M = now;
        }
    }
    return m;
}
ll pow_ll(ll x, ll n) {
    if (n == 0) return 1;
    if (n % 2) {
        return pow_ll(x, n - 1) * x;
    }
    else {
        ll tmp = pow_ll(x, n / 2);
        return tmp * tmp;
    }
}
ll keta(ll num, ll arity) {
    ll ret = 0;
    while (num) {
        num /= arity;
        ret++;
    }
    return ret;
}
// k進数で見た時のi桁目の数を返す (一番下は0桁目)
ll keta_num(ll num, ll i, ll k) {
    return (num / pow_ll(k, i)) % k;
}
ll ceil(ll n, ll m) {
    // n > 0, m > 0
    ll ret = n / m;
    if (n % m) ret++;
    return ret;
}
void compress(vector<ll>& v) {
    // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] 
    vector<ll> u = v;
    sort(u.begin(), u.end());
    u.erase(unique(u.begin(),u.end()),u.end());
    map<ll, ll> mp;
    for (int i = 0; i < u.size(); i++) {
        mp[u[i]] = i;
    }
    for (int i = 0; i < v.size(); i++) {
        v[i] = mp[v[i]];
    }
}
// N以下の素数を全て返す
vector<ll> Eratosthenes( const ll N )
{
    vector<bool> is_prime( N + 1 );
    for( ll i = 0; i <= N; i++ )
    {
        is_prime[ i ] = true;
    }
    vector<ll> P;
    for( ll i = 2; i <= N; i++ )
    {
        if( is_prime[ i ] )
        {
            for( ll j = 2 * i; j <= N; j += i )
            {
                is_prime[ j ] = false;
            }
            P.emplace_back( i );
        }
    }
    return P;
}
vector<pair<ll, ll> > prime_factorize(ll N) {
    vector<pair<ll, ll> > res;
    for (ll a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        ll ex = 0; // 指数
        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }
        // その結果を push
        res.push_back({a, ex});
    }
    // 最後に残った数について
    if (N != 1) res.push_back({N, 1});
    return res;
}
#pragma endregion
// 3点の座標から面積を求める
double area(P a, P b, P c) {
    return abs((b.first - a.first) * (c.second - a.second) - (b.second - a.second) * (c.first - a.first)) / 2.0;
}
ll dx[] = {1, -1, 0, 0};
ll dy[] = {0, 0, 1, -1};
int main() {
    ll x1, y1, x2, y2, x3, y3;
    cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
    double max_area = 0;
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            for (int k = 0; k < 4; k++) {
                ll nx1 = x1 + dx[i];
                ll ny1 = y1 + dy[i];
                ll nx2 = x2 + dx[j];
                ll ny2 = y2 + dy[j];
                ll nx3 = x3 + dx[k];
                ll ny3 = y3 + dy[k];
                max_area = max(max_area, area(P(nx1, ny1), P(nx2, ny2), P(nx3, ny3)));
            }
        }
    }
    cout << fixed << setprecision(10) << max_area << endl;
    return 0;
}
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