Submission #883315

#	Time	Username	Problem	Language	Result	Execution time	Memory
883315		phoenix0423	Skyscraper (JOI16_skyscraper)	C++17	100 / 100	174 ms	145160 KiB

skyscraper

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
#define fastio ios::sync_with_stdio(false), cin.tie(0)
// #pragma GCC optimize("Ofast")
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define REP1(i, n) FOR(i, 0, n + 1)
#define SZ(x) (int)(x).size()
#define pb push_back
#define pf push_front
#define eb emplace_back
#define f first
#define s second
#define lowbit(x) x&-x
#define ckmin(a, b) a = min(a, b)
#define ckmax(a, b) a = max(a, b)
#define int long long
const int INF = 1e18;
const int maxn = 1000 + 5;
const int N = 1e9 + 7;
ll dp[105][105][2005][3];

signed main(void){
	fastio;
	int n, L;
	cin>>n>>L;
	vector<int> a(n);
	for(int i = 0; i < n; i++) cin>>a[i];
	if(n == 1){
		cout<<1<<"\n";
		return 0;
	}
	sort(a.begin(), a.end());
	a.pb(a[n - 1]);
	dp[0][0][0][0] = 1;
	for(int i = 0; i < n; i++){
		for(int j = 0; j <= i; j++){
			for(int k = 0; k <= L; k++){
				for(int l = 0; l <= 2; l++){
					dp[i][j][k][l] %= N;
					if(k + (2 * j - 2 - l) * (a[i + 1] - a[i]) > L) continue;
					if(j) (dp[i + 1][j][k + (2 * j - l) * (a[i + 1] - a[i])][l] += dp[i][j][k][l] * (2 * j - l)) %= N;
					(dp[i + 1][j + 1][k + (2 * j + 2 - l) * (a[i + 1] - a[i])][l] += dp[i][j][k][l] * (j + 1 - l)) %= N;
					if(j > 1) (dp[i + 1][j - 1][k + (2 * j - 2 - l) * (a[i + 1] - a[i])][l] += dp[i][j][k][l] * (j - 1)) %= N;
					if(l != 2){
						(dp[i + 1][j][k + (2 * j - (l + 1)) * (a[i + 1] - a[i])][l + 1] += dp[i][j][k][l] * (2 - l)) %= N;
						(dp[i + 1][j + 1][k + (2 * j + 2 - (l + 1)) * (a[i + 1] - a[i])][l + 1] += dp[i][j][k][l] * (2 - l)) %= N;
					}
					// if(dp[i][j][k][l]) cout<<i<<" "<<j<<" "<<k<<" "<<l<<" : "<<dp[i][j][k][l]<<"\n";
				}
			}
		}
	}
	ll ans = 0;
	// for(int i = 0; i <= L; i++) (ans += dp[n - 1][2][i][2]) %= N;
	// for(int i = 0; i <= L; i++) (ans += dp[n - 1][1][i][1]) %= N;
	for(int i = 0; i <= L; i++) (ans += dp[n][1][i][2]) %= N;
	cout<<ans<<"\n";

}

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