Submission #1020830

#	Time	Username	Problem	Language	Result	Execution time	Memory
1020830		flappybird	Boat (APIO16_boat)	C++17	27 / 100	1374 ms	8276 KiB

boat

#include <bits/stdc++.h>
#include <cassert>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
#define MAX 510
#define MAXQ 101010
#define INF 1000000100
#define bb ' '
#define ln '\n'
#define Ln '\n'
#define MOD 1000000007
int A[MAX];
int B[MAX];
int sz[MAX];
ll C[MAX * 2][MAX];
ll dp[MAX][MAX * 2];
ll mpow(ll x, ll y = MOD - 2) {
	if (!y) return 1;
	ll res = mpow(x, y >> 1);
	res *= res;
	res %= MOD;
	if (y & 1) res = (res * x) % MOD;
	return res;
}
ll rev[MAX * 2];
signed main() {
	ios::sync_with_stdio(false), cin.tie(0);
	int N;
	cin >> N;
	int i, j, k;
	vector<int> v;
	for (i = 1; i <= N; i++) {
		cin >> A[i] >> B[i];
		B[i]++;
		v.push_back(A[i]);
		v.push_back(B[i]);
	}
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
	for (i = 1; i <= N; i++) {
		A[i] = lower_bound(v.begin(), v.end(), A[i]) - v.begin();
		B[i] = lower_bound(v.begin(), v.end(), B[i]) - v.begin();
	}
	for (i = 1; i <= N * 2; i++) rev[i] = mpow(i);
	int X = v.size() - 1;
	for (i = 0; i < X; i++) {
		sz[i] = v[i + 1] - v[i];
		for (j = 1; j <= N; j++) {
			ll c1, c2;
			c1 = 1;
			c2 = sz[i];
			C[i][j] += c1 * c2;
			C[i][j] %= MOD;
			for (k = 0; k < j - 1; k++) {
				c1 *= j - 1 - k;
				c1 %= MOD;
				c1 *= rev[k + 1];
				c1 %= MOD;
				c2 *= sz[i] - (k + 1);
				c2 %= MOD;
				c2 *= rev[k + 2];
				c2 %= MOD;
				C[i][j] += c1 * c2;
				C[i][j] %= MOD;
			}
		}
	}
	for (i = 1; i <= N; i++) {
		int cnt = 1;
		if (A[i] == 0) {
			for (j = 1; j < i; j++) if (A[j] == 0) cnt++;
			dp[i][0] += C[0][cnt];
			dp[i][0] %= MOD;
		}
		for (j = 1; j < X; j++) {
			dp[i][j] = dp[i][j - 1];
			if (B[i] <= j) continue;
			if (j < A[i]) continue;
			cnt = 1;
			for (k = i - 1; k >= 1; k--) {
				dp[i][j] += C[j][cnt] * dp[k][j - 1];
				dp[i][j] %= MOD;
				if (A[k] <= j && j < B[k]) cnt++;
			}
			dp[i][j] += C[j][cnt];
			dp[i][j] %= MOD;
		}
	}
	ll ans = 0;
	for (i = 1; i <= N; i++) ans = (ans + dp[i][X - 1]) % MOD;
	cout << ans;
}

• Subtask 1: 0 / 9
• Subtask 2: 0 / 22
• Subtask 3: 27 / 27
• Subtask 4: 0 / 42