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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
#208481#6677. Puzzle: SashiganePetroTarnavskyi#WA 1ms3492kbC++171.7kb2023-10-09 17:37:272023-10-09 17:37:28

Judging History

你现在查看的是最新测评结果

  • [2023-10-09 17:37:28]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:3492kb
  • [2023-10-09 17:37:27]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

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

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

const int N = 1 << 19;

struct Point
{
	int x, y;
	Point() {}
	Point(int _x, int _y): x(_x), y(_y) {}
	Point operator-(const Point& p) const
	{
		return {x - p.x, y - p.y};
	}
	LL operator*(const Point& p) const
	{
		return (LL)x * p.y - (LL)y * p.x;
	}
} p[N];

LL area(int i, int j, int k)
{
	LL s = (p[j] - p[i]) * (p[k] - p[i]);
	assert(s >= 0);
	return s;
}

void solve()
{
	int n, k;
	cin >> n >> k;
	FOR(i, 0, n)
	{
		cin >> p[i].x >> p[i].y;
		p[n + i] = p[i];
		p[2 * n + i] = p[i];
	}
	LL sbc = 0;
	FOR(i, 2, k + 1)
		sbc += area(0, 1, i);
	int ptr = k;
	LL res = 0;
	FOR(i, 0, n)
	{
		while (ptr < i + k || (ptr + 1 < n + i && area(i, i + k, ptr + 1) > area(i, i + k, ptr)))
			ptr++;
		FOR(j, i + k, n + i + 1)
			res = max(res, sbc + area(i, i + k, j));
		if (i == 2)
		{
			cerr << i << " " << i + k << " " << ptr << " " << area(i, i + k, ptr) << " " << sbc << endl;
		}
		res = max(res, sbc + area(i, i + k, ptr));
		sbc += area(i, i + k, i + k + 1) - area(i, i + 1, i + k + 1);
	}
	cout << res / 2;
	if (res % 2 == 1)
		cout << ".5";
	cout << "\n";
}

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(15);
	int t;
	cin >> t;
	while (t--)
		solve();
	return 0;	
}



詳細信息

Test #1:

score: 0
Wrong Answer
time: 1ms
memory: 3492kb

input:

5 3 4

output:

0
0
0
0
0

result:

wrong answer YES or NO expected in answer, but 0 found. (test case 1)