Submission #933391

#	Time	Username	Problem	Language	Result	Execution time	Memory
933391		parlimoos	Skyscraper (JOI16_skyscraper)	C++14	100 / 100	43 ms	50772 KiB

skyscraper

//Be Name KHODA
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
#define pb push_back
#define pp pop_back
#define lb lower_bound
#define ub upper_bound
#define cl clear
#define bg begin
#define arr(x) array<int , x>
#define endl '\n'

const int MOD = int(1e9) + 7;

int n , l;
vector<int> a;
ll dp[101][101][1001][3];

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> l;
    a.pb(0);
    for(int i = 0 ; i < n ; i++){
        int d;
        cin >> d;
        a.pb(d);
    }
    a.pb(10000);
    if(n == 1){
        cout << 1;
        return 0;
    }
    sort(a.bg() , a.end());
    dp[0][0][0][0] = 1;
    for(int i = 1 ; i <= n ; i++){
        for(int j = 1 ; j <= i ; j++){
            for(int k = 0 ; k <= l ; k++){
                for(int m = 0 ; m < 3 ; m++){
                    int d = (2 * j - m) * (a[i + 1] - a[i]);
                    if(d > k or i + j + 1 - m > n) continue;
                    dp[i][j][k][m] += dp[i - 1][j - 1][k - d][m];
                    if(m > 0) dp[i][j][k][m] += (3 - m) * dp[i - 1][j - 1][k - d][m - 1];

                    dp[i][j][k][m] += (2 * j - m) * dp[i - 1][j][k - d][m];
                    
                    if(m == 1) dp[i][j][k][m] += (2 * j) * dp[i - 1][j][k - d][m - 1];
                    else if(m == 2){
                        if(i == n) dp[i][j][k][m] += dp[i - 1][j][k - d][m - 1];
                        else if(j > 1) dp[i][j][k][m] += (j - 1) * dp[i - 1][j][k - d][m - 1];
                    }

                    if(m == 2){
                        if(i == n) dp[i][j][k][m] += dp[i - 1][j + 1][k - d][m];
                        else dp[i][j][k][m] += (j - 1) * j * dp[i - 1][j + 1][k - d][m];
                    }else if(m == 1) dp[i][j][k][m] += j * j * dp[i - 1][j + 1][k - d][m];
                    else dp[i][j][k][m] += j * (j + 1) * dp[i - 1][j + 1][k - d][m];

                    dp[i][j][k][m] %= MOD;
                }
            }
        }
    }
    ll o = 0;
    for(int i = 0 ; i <= l ; i++) o += dp[n][1][i][2];
    cout << o % MOD;
}

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