제출 #676565

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
676565		QwertyPi	생물 실험 (IZhO13_expgorl)	C++14	100 / 100	2 ms	724 KiB

expgorl

#include <bits/stdc++.h>
#define int long long
using namespace std;

struct point{
	int x, y;
};

int dx[256], dy[256];

int dis(point A, point B){
	return (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y);
}

long long low = 1LL << 60, high = -(1LL << 60);
point L, G, D;
void check(point X){
	int d = dis(X, D);
	low = min(low, d);
	high = max(high, d); 
}

int K; 
int lim(int x){
	return max(min(x, K - 1), 0LL);
}

int32_t main(){
	cin >> K;
	string s; cin >> s;
	dx['L'] = -1; dy['L'] =  0;
	dx['R'] =  1; dy['R'] =  0;
	dx['F'] =  0; dy['F'] =  1;
	dx['B'] =  0; dy['B'] = -1;
	dx['I'] =  0; dy['I'] =  0;
	int x, y; cin >> x >> y; L = {x, y}; cin >> x >> y; G = {x, y};
	D = {L.x - G.x, L.y - G.y};
	
	
	point X {0, 0};
	vector<point> pts;
	for(auto c : s){
		pts.push_back(X);
		X.x += dx[c]; X.y += dy[c];
	}
	
	int cx = X.x, cy = X.y;
	check({cx * K, cy * K});
	for(auto p : pts){
		check(p); check({p.x + cx * (K - 1), p.y + cy * (K - 1)});
	}
	
	if(cx * cx + cy * cy != 0){
		for(auto p : pts){
			int a = cx * cx + cy * cy, b = -2LL * ((D.x - p.x) * cx + (D.y - p.y) * cy), c = (D.x - p.x) * (D.x - p.x) + (D.y - p.y) * (D.y - p.y);
			int opt = -b / a / 2;
			
			auto check_2 = [&p, &cx, &cy](int x){
				check({p.x + lim(x) * cx, p.y + lim(x) * cy});
			};
			
			check_2(opt - 1); check_2(opt); check_2(opt + 1);
		}
	}
	cout << fixed << setprecision(12) << sqrt(low) << ' ' << sqrt(high) << endl; 
}

컴파일 시 표준 에러 (stderr) 메시지

expgorl.cpp: In function 'int32_t main()':
expgorl.cpp:44:13: warning: array subscript has type 'char' [-Wchar-subscripts]
   44 |   X.x += dx[c]; X.y += dy[c];
      |             ^
expgorl.cpp:44:27: warning: array subscript has type 'char' [-Wchar-subscripts]
   44 |   X.x += dx[c]; X.y += dy[c];
      |                           ^
expgorl.cpp:55:81: warning: unused variable 'c' [-Wunused-variable]
   55 |    int a = cx * cx + cy * cy, b = -2LL * ((D.x - p.x) * cx + (D.y - p.y) * cy), c = (D.x - p.x) * (D.x - p.x) + (D.y - p.y) * (D.y - p.y);
      |                                                                                 ^

• Subtask 1: 100 / 100