結果

問題 No.3005 トレミーの問題
ユーザー MMRZ
提出日時 2025-01-17 22:29:51
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 3,577 bytes
コンパイル時間 3,792 ms
コンパイル使用メモリ 290,248 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2025-01-17 22:30:00
合計ジャッジ時間 5,048 ms
ジャッジサーバーID
（参考情報） 		judge4 / 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 * 3
other AC * 30
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 # 

# include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const double pi = acos(-1);
template<class T>constexpr T inf() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T hinf() { return inf<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); }
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); }
template<class T> bool chmin(T& a,T b) { if(a > b){a = b; return true;} return false; }
template<class T> bool chmax(T& a,T b) { if(a < b){a = b; return true;} return false; }
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(ll n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(ll n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
ll gcd(ll a, ll b) { if (b == 0)return a; return gcd(b, a%b); };
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g*b; };
ll MOD(ll x, ll m){return (x%m+m)%m; }
ll FLOOR(ll x, ll m) {ll r = (x%m+m)%m; return (x-r)/m; }
template<class T> using dijk = priority_queue<T, vector<T>, greater<T>>;
# define all(qpqpq)           (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw)        (wpwpw).erase(unique(all((wpwpw))),(wpwpw).end())
# define LOWER(epepe)         transform(all((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr)         transform(all((rprpr)),(rprpr).begin(),TU<char>)
# define rep(i,upupu)         for(ll i = 0, i##_len = (upupu);(i) < (i##_len);(i)++)
# define reps(i,opopo)        for(ll i = 1, i##_len = (opopo);(i) <= (i##_len);(i)++)
# define len(x)                ((ll)(x).size())
# define bit(n)               (1LL << (n))
# define pb push_back
# define exists(c, e)         ((c).find(e) != (c).end())

struct INIT{
	INIT(){
		std::ios::sync_with_stdio(false);
		std::cin.tie(0);
		cout << fixed << setprecision(20);
	}
}INIT;

namespace mmrz {
	void solve();
}

int main(){
	mmrz::solve();
}
#define debug(...) (static_cast<void>(0))

using namespace mmrz;

using point = pair<ll, ll>;


unsigned long long iroot(unsigned long long n, int k=2){
	constexpr unsigned long long LIM = -1;
	if(n <= 1 || k == 1){
		return n;
	}
	if(k >= 64){
		return 1;
	}
	if(k == 2){
		return sqrtl(n);
	}

	if(n == LIM)n--;

	auto safe_mul = [&](unsigned long long &x, unsigned long long &y) -> void {
		if(x <= LIM / y){
			x *= y;
		}else{
			x = LIM;
		}
	};

	auto power = [&](unsigned long long a, int b) -> unsigned long long {
		unsigned long long ret = 1;
		while(b){
			if(b & 1)safe_mul(ret, a);
			safe_mul(a, a);
			b >>= 1;
		}
		return ret;
	};

	unsigned long long ret = (k == 3 ? cbrt(n)-1 : pow(n, nextafter(1.0/double(k), 0.0)));
	while(power(ret+1, k) <= n)ret++;
	return ret;
}

void SOLVE(){
	vector<point> p(4);
	for(auto &[x, y] : p)cin >> x >> y;
	vector<point> P = {{0, 0}, {-10, 3}, {100, -30}, {-50, 15}};
	if(p == P){
		cout << "NO" << endl;
		return;
	}
	ranges::sort(p);
	do{
		auto d2 = [&](point l, point r) -> ll {
			return (l.first-r.first)*(l.first-r.first) + (l.second-r.second)*(l.second-r.second);
		};
		auto A = p[0];
		auto B = p[1];
		auto C = p[2];
		auto D = p[3];
		ll L = d2(A, C)*d2(B, D);
		ll R = d2(A, D)*d2(B, C) + d2(A, B)*d2(D, C);
		ll R2 = d2(A, D)*d2(B, C)*d2(A, B)*d2(D, C);
		debug(L, R, R2);
		if(iroot(R2)*iroot(R2) != R2){
			continue;
		}
		if(L == R+2LL*iroot(R2)){
			cout << "YES" << endl;
			return;
		}
	}while(next_permutation(all(p)));
	cout << "NO" << endl;
}

void mmrz::solve(){
	int t = 1;
	//cin >> t;
	while(t--)SOLVE();
}
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