Submission #468134

#	Time	Username	Problem	Language	Result	Execution time	Memory
468134		sinamhdv	Skyscraper (JOI16_skyscraper)	C++11	100 / 100	149 ms	2924 KiB

skyscraper

// JOI16_skyscraper
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1000 * 1000 * 1000 + 7;
const int INF = 1e9 + 100;
const ll LINF = 1e18 + 100;

#ifdef DEBUG
#define dbg(x) cout << #x << " = " << (x) << endl << flush;
#define dbgr(s, f) { cout << #s << ": "; for (auto _ = (s); _ != (f); _++) cout << *_ << ' '; cout << endl << flush; }
#else
#define dbg(x) ;
#define dbgr(s, f) ;
#endif
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define pb push_back
#define fr first
#define sc second
#define all(x) (x).begin(), (x).end()
#define endl '\n'

const int MAXN = 110;
const int MAXL = 1010;

int n, L, a[MAXN];
int dp[2][MAXN][MAXL][3];

inline void relax(int i, int j, int k, int l, ll val)
{
	if (k < 0 || k >= MAXL) return;
	dp[i][j][k][l] = (dp[i][j][k][l] + val) % mod;
}

int32_t main(void)
{
	fast_io;
	cin >> n >> L;
	if (n == 1) return cout << "1\n", 0;

	FOR(i, 1, n + 1) cin >> a[i];
	sort(a + 1, a + n + 1);

	// base
	FOR(k, 0, MAXL)
	{
		dp[1][1][k][0] = 1;
		dp[1][1][k][1] = 1;
	}

	// dp
	FOR(i, 1, n)
	{
		int p = i % 2, c = 1 - p;
		int dif = a[i + 1] - a[i];
		memset(dp[c], 0, sizeof(dp[c]));

		FOR(j, 1, i + 1)
		{
			FOR(k, 0, MAXL)
			{
				relax(c, j + 1, k + 2 * j * dif, 0, (j + 1) * (ll)dp[p][j][k][0]);
				relax(c, j, k + 2 * j * dif, 0, 2 * j * (ll)dp[p][j][k][0]);
				relax(c, j - 1, k + 2 * j * dif, 0, (j - 1) * (ll)dp[p][j][k][0]);

				
				

				relax(c, j + 1, k + (2 * j - 1) * dif, 1, j * (ll)dp[p][j][k][1]);
				relax(c, j + 1, k + 2 * j * dif, 1, dp[p][j][k][0]);
				
				relax(c, j, k + (2 * j - 1) * dif, 1, (2 * j - 1) * (ll)dp[p][j][k][1]);
				relax(c, j, k + 2 * j * dif, 1, dp[p][j][k][0]);
				
				relax(c, j - 1, k + (2 * j - 1) * dif, 1, (j - 1) * (ll)dp[p][j][k][1]);


				

				relax(c, j + 1, k + (2 * j - 2) * dif, 2, (j - 1) * (ll)dp[p][j][k][2]);
				relax(c, j + 1, k + (2 * j - 1) * dif, 2, 2ll * dp[p][j][k][1]);
				
				relax(c, j, k + (2 * j - 2) * dif, 2, (2 * j - 2) * (ll)dp[p][j][k][2]);
				relax(c, j, k + (2 * j - 1) * dif, 2, 2ll * dp[p][j][k][1]);
				
				relax(c, j - 1, k + (2 * j - 2) * dif, 2, (j - 1) * (ll)dp[p][j][k][2]);
			}
		}

		if (i + 1 < n)
		{
			FOR(k, 0, MAXL) dp[c][1][k][2] = 0;
		}
/*
		if (i == 2)
		{
			FOR(j, 1, i + 2) FOR(l, 0, 3)
			{
				cout << j << ' ' << l << endl;
				FOR(k, 0, 10) cout << dp[c][j][k][l] << ' ';
				cout << endl;
			}
		}
*/
	}

	int ans = dp[n % 2][1][L][2];
	ans = (mod + ans % mod) % mod;
	cout << ans << endl;

	return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3: 80 / 80