Submission #569170

#	Time	Username	Problem	Language	Result	Execution time	Memory
569170		rawoti6303	NoM (RMI21_nom)	C++14	100 / 100	71 ms	27936 KiB

Main

#include <bits/stdc++.h>

#define INF 1000000005
#define INFL 1000000000000000005ll
#define EPS 1e-6
#define pb push_back
#define pause system("pause")
#define exit exit(0)
#define endl '\n'

using namespace std;
using ull = unsigned long long;
using ll = long long;

typedef pair<ll, ll> pii;
const int N = 2005, MOD = 1e9 + 7;

int n, m, f[N], c[N][N], dp[N][N];

inline int add(int a, int b) {
	return a + b < MOD ? (a + b) : (a + b - MOD);
}

inline int mul(int a, int b) {
	return ((ll)a * b) % MOD;
}

inline void upd(int &a, int b) {
	a = add(a, b);
}

inline int bpow(int a, int b) {
	int ans = 1;
	while (b != 0) {
		if (b & 1)
			ans = mul(ans, a);
		a = mul(a, a), b >>= 1;
	}

	return ans;
}

int fact(int n) {
	int ans = 1;
	for (int i = 2; i <= n; ++i) {
		ans = mul(ans, i);
	}

	return ans;
}

int naive(int n, int m) {
	vector<int> p(2 * n);
	for (int i = 0; i < 2 * n; ++i) {
		p[i] = i + 1;
	}

	int ans = 0;
	do {
		bool ok = 1;
		for (int i = 0; ok && i < 2 * n; ++i) {
			for (int j = i + 1; j < 2 * n; ++j) {
				if (abs(p[i] - p[j]) == n) {
					ok &= ((j - i) % m) != 0;
					break;
				}
			}
		}

		ans += ok;
	} while (next_permutation(p.begin(), p.end()));
	return ans;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n >> m;
	//cout << naive(n, m) << endl;
	for (int i = 0; i <= n; ++i) {
		c[i][0] = 1;
		for (int j = 1; j <= i; ++j) {
			c[i][j] = add(c[i - 1][j - 1], c[i - 1][j]);
		}
	}

	for (int i = 0; i < 2 * n; ++i) {
		++f[i % m];
	}

	int sm = f[0];
	dp[0][f[0]] = mul(c[n][f[0]], bpow(2, f[0]));
	for (int i = 0; i < m - 1; ++i) {
		for (int j = 0; j <= sm; ++j) {
			for (int k = 0; k <= j && k <= f[i + 1]; ++k) {
				if (j - k + f[i + 1] - k > n || dp[i][j] == 0)
					continue;

				upd(dp[i + 1][j - k + f[i + 1] - k],
					mul(
						dp[i][j],
						mul(
							c[j][k],
							mul(
								c[n - (((sm - j) / 2) + j)][f[i + 1] - k],
								bpow(2, f[i + 1] - k)
							)
						)
					)
				);
			}
		}

		sm += f[i + 1];
	}

	int ans = dp[m - 1][0];
	for (int i = 0; i < m; ++i) {
		ans = mul(ans, fact(f[i]));
	}

	cout << ans << endl;
	//pause;
	return 0;
}

• Subtask 1: 9 / 9
• Subtask 2: 12 / 12
• Subtask 3: 13 / 13
• Subtask 4: 18 / 18
• Subtask 5: 48 / 48