Submission #1193926

#	Time	Username	Problem	Language	Result	Execution time	Memory
1193926		Sir_Ahmed_Imran	Skyscraper (JOI16_skyscraper)	C++17	100 / 100	107 ms	48488 KiB

skyscraper

            //    01001100 01001111 01010100 01000001    \\
            //                                           \\
            //                ╦  ╔═╗╔╦╗╔═╗               \\
            //                ║  ║ ║ ║ ╠═╣               \\
            //                ╩═╝╚═╝ ╩ ╩ ╩               \\
            //                                           \\
            //    01001100 01001111 01010100 01000001    \\

#include <bits/stdc++.h>
using namespace std;
#define N 101
#define K 1001
#define nl '\n'
#define ff first
#define ss second
#define add insert
#define ll long long
#define ld long double
#define terminator main
#define pll pair<ll,ll>
#define append push_back
#define pii pair<int,int>
#define all(x) (x).begin(),(x).end()
#define L0TA ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

const int M = 1e9 + 7;

int add(int x, int y){
    return (x + y) % M;
}

int mul(int x, int y){
    return (1ll * x * y) % M;
}

int a[N];
int dp[N][N][K][3];

void solve(){
    int n, L, val, ans;
    cin >> n >> L;
    for(int i = 1; i <= n; i++)
        cin >> a[i];
    dp[1][1][0][0] = 1;
    dp[1][1][0][1] = 2;
    dp[1][1][0][2] = 1;
    sort(a + 1, a + n + 1);
    for(int i = 2; i <= n; i++){
        for(int j = 1; j < i; j++){
            for(int l = 0; l <= L; l++){
                for(int k = 0; k < 3; k++){
                    val = (a[i] - a[i - 1]) * (2 * j - k) + l;
                    if(val > L) continue;
                    dp[i][j - 1][val][k] = add(dp[i][j - 1][val][k], mul(dp[i - 1][j][l][k], j - 1));
                    dp[i][j][val][k] = add(dp[i][j][val][k], mul(dp[i - 1][j][l][k], 2 * j - k));
                    dp[i][j + 1][val][k] = add(dp[i][j + 1][val][k], mul(dp[i - 1][j][l][k], j + 1 - k));
                    if(k > 1) continue;
                    dp[i][j][val][k + 1] = add(dp[i][j][val][k + 1], mul(dp[i - 1][j][l][k], 2 - k));
                    dp[i][j + 1][val][k + 1] = add(dp[i][j + 1][val][k + 1], mul(dp[i - 1][j][l][k], 2 - k));
                }
            }
        }
    }
    for(int i = ans = 0; i <= L; i++)
        ans = add(ans, dp[n][1][i][2]);
    cout << ans;
}

int terminator(){
    L0TA;
    solve();
    return 0;
}

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