結果

問題 No.2515 Similar Triangles
ユーザー baluteshih
提出日時 2023-10-27 21:24:52
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 1,918 bytes
コンパイル時間 1,908 ms
コンパイル使用メモリ 196,360 KB
最終ジャッジ日時 2025-02-17 14:28:12
ジャッジサーバーID
(参考情報)
judge1 / judge2
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 * 7
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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;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define X first
#define Y second
#define SZ(a) ((int)a.size())
#define ALL(v) v.begin(), v.end()
#define pb push_back
typedef pair<double, double> pdd;
typedef pair<pll, pll> Line;
pll operator+(pll a, pll b)
{ return pll(a.X + b.X, a.Y + b.Y); }
pll operator-(pll a, pll b)
{ return pll(a.X - b.X, a.Y - b.Y); }
pll operator*(pll a, ll b)
{ return pll(a.X * b, a.Y * b); }
pll operator/(pll a, ll b)
{ return pll(a.X / b, a.Y / b); }
pdd operator/(pll a, double b)
{ return pdd(a.X / b, a.Y / b); }
ll dot(pll a, pll b)
{ return a.X * b.X + a.Y * b.Y; }
ll cross(pll a, pll b)
{ return a.X * b.Y - a.Y * b.X; }
ll abs2(pll a)
{ return dot(a, a); }
int sign(ll a)
{ return a == 0 ? 0 : a > 0 ? 1 : -1; }
int ori(pll a, pll b, pll c)
{ return sign(cross(b - a, c - a)); }
bool collinearity(pll p1, pll p2, pll p3)
{ return sign(cross(p1 - p3, p2 - p3)) == 0; }
bool btw(pll p1, pll p2, pll p3) {
if (!collinearity(p1, p2, p3)) return 0;
return sign(dot(p1 - p3, p2 - p3)) <= 0;
}
bool seg_intersect(pll p1, pll p2, pll p3, pll p4) {
int a123 = ori(p1, p2, p3);
int a124 = ori(p1, p2, p4);
int a341 = ori(p3, p4, p1);
int a342 = ori(p3, p4, p2);
if (a123 == 0 && a124 == 0)
return btw(p1, p2, p3) || btw(p1, p2, p4) ||
btw(p3, p4, p1) || btw(p3, p4, p2);
return a123 * a124 <= 0 && a341 * a342 <= 0;
}
pdd intersect(pll p1, pll p2, pll p3, pll p4) {
ll a123 = cross(p2 - p1, p3 - p1);
ll a124 = cross(p2 - p1, p4 - p1);
return (p4 * a123 - p3 * a124) / double(a123 - a124); // C^3 / C^2
}
pll perp(pll p1)
{ return pll(-p1.Y, p1.X); }
int main() {
ios::sync_with_stdio(0), cin.tie(0);
pll a, b, c;
cin >> a.X >> a.Y >> b.X >> b.Y >> c.X, c.Y = 0;
if (abs(cross(a, c)) == abs(cross(b, c)))
cout << "Yes\n";
else
cout << "No\n";
}
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