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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#102622#3189. Finding LinesPetroTarnavskyi#TL 2ms3464kbC++171.2kb2023-05-03 15:13:422023-05-03 15:14:01

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-05-03 15:14:01]
  • 评测
  • 测评结果:TL
  • 用时:2ms
  • 内存:3464kb
  • [2023-05-03 15:13:42]
  • 提交

answer

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

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

struct point
{
	LL x, y;
	
	point(): x(0), y(0) {}
	point(LL _x, LL _y): x(_x), y(_y) {}
	
	point operator -(const point p) const
	{
		return point(x - p.x, y - p.y);
	}
	long long operator *(const point p) const 
	{
		return x * p.y - y * p.x;
	}
};

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	srand(47);
	int n, p;
	cin >> n >> p;
	if (n < 3)
	{
		cout << "possible\n";
		return 0;
	}
	vector<point> v(n);
	FOR(i, 0, n) cin >> v[i].x >> v[i].y;
	
	while(double(clock()) / CLOCKS_PER_SEC < 3.74)
	{
		int i = rand() % n;
		int j = rand() % n;
		if (i == j) continue;
		int cnt = 0;
		FOR(k, 0, n)
		{
			if ((v[k] - v[i]) * (v[j] - v[i]) == 0)
				cnt++;
		}
		// cnt / n >= p /100
		if (100 * cnt >= n * p)
		{
			cout << "possible\n";
			return 0;
		}
	}
	cout << "impossible\n";
		
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 2ms
memory: 3464kb

input:

5
55
0 0
10 10
10 0
0 10
3 3

output:

possible

result:

ok single line: 'possible'

Test #2:

score: -100
Time Limit Exceeded

input:

5
45
0 0
10 10
10 0
0 10
3 4

output:


result: