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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#704403#5414. Stop, Yesterday Please No MoreKanateWA 0ms3660kbC++172.0kb2024-11-02 19:54:542024-11-02 19:54:55

Judging History

你现在查看的是最新测评结果

  • [2024-11-02 19:54:55]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3660kb
  • [2024-11-02 19:54:54]
  • 提交

answer

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

const int M = 2e3 + 7;
const int N = 1e6 + 7;

int T;
int n, m, k;
int ans;
string s;

int a[M][M];
int fx[4] = {0, 0, 1, -1};
int fy[4] = {1, -1, 0, 0};

int get(char s)
{
	if (s == 'U')
		return 3;
	if (s == 'L')
		return 1;
	if (s == 'D')
		return 2;
	return 0;
}

int pre(int x1, int y1, int x2, int y2)
{
	int ans = a[x2][y2] + a[x1 - 1][y1 - 1] - a[x1 - 1][y2] - a[x2][y1 - 1];
	return ans;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> T;
	while (T-- > 0)
	{
		cin >> n >> m >> k;
		for (int i = 1; i <= 2 * n; i++)
			for (int j = 1; j <= 2 * m; j++)
				a[i][j] = 0;
		cin >> s;
		bool flag = 0;
		ans = 0;
		int dx = n, dy = m;
		int mxx = n, mix = 1, mxy = m, miy = 1;
		a[n][m] = 1;
		for (auto v : s)
		{
			int nx = get(v);
			dx += fx[nx];
			dy += fy[nx];
			if (dx <= 0 || dx > 2 * n - 1 || dy <= 0 || dy > 2 * m - 1)
			{
				if (k == 0)
					cout << n * m << endl;
				else
					cout << 0 << endl;
				flag = 1;
				break;
				
			}a[dx][dy] = 1;
				mxx = min(mxx, 2 * n - dx);
				mix = max(mix, n - dx + 1);
				mxy = min(mxy, 2 * m - dy);
				miy = max(miy, m - dy + 1);
			if (flag == 1)
				continue;
			if (mix > mxx || miy > mxy)
			{
				if (k == 0)
					cout << n * m << endl;
				else
					cout << 0 << endl;
				flag = 1;
			}
			if (flag == 1)
				continue;
			for (int i = 1; i <= n * 2; i++)
				for (int j = 1; j <= m * 2; j++)
					a[i][j] += a[i][j - 1];
			for (int j = 1; j <= m * 2; j++)
				for (int i = 1; i <= n * 2; i++)
					a[i][j] += a[i - 1][j];
			int now = (mxx - mix + 1) * (mxy - miy + 1);
			for (int i = 1; i <= n; i++)
				for (int j = 1; j <= m; j++)
				{
					int x1 = mix - i + n;
					int x2 = mxx - i + n;
					int y1 = miy - j + m;
					int y2 = mxy - j + m;
					if (now - pre(x1, y1, x2, y2) == k)
					{
						ans++;
					}
				}
			cout << ans << endl;
		}
	}
}

Details

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Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 3660kb

input:

3
4 5 3
ULDDRR
4 5 0
UUUUUUU
4 5 10
UUUUUUU

output:

0
0
0
0
1
1
0
0
0
20
0
0
0
0

result:

wrong answer 1st numbers differ - expected: '2', found: '0'