Submission #23300

#	Time	Username	Problem	Language	Result	Execution time	Memory
23300		duongthoi1999	Skyscraper (JOI16_skyscraper)	C++14	100 / 100	76 ms	4768 KiB

skyscraper

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef long long ll;
typedef long double ld;
typedef pair<ll, int> ii;
const int mod = (int) 1e9 + 7;
const ll inf = 1LL << 60;
const int maxn = (int) 105;
const ld eps = 1e-9;

void add(int &a, ll b) {
    b %= mod;
    a += b;
    while (a >= mod) a -= mod;
}

int n, L, a[maxn];
int f[2][maxn][maxn * 10][3];

void trau() {
    int cnt = 0;
    do {
        int res = 0;
        rep(i, 1, n) res += abs(a[i] - a[i + 1]);
        if(res <= L) cnt ++;
    } while (next_permutation(a + 1, a + 1 + n));
    cout << cnt << endl;
}

int main() {
    //freopen("test.txt", "r", stdin);
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> L;
    rep(i, 1, n + 1) cin >> a[i];
    sort(a + 1, a + 1 + n);
    a[n + 1] = 12345;
    if(a[2] - a[1] <= L) f[1][1][a[2] - a[1]][1] = 2;
    if(2 * (a[2] - a[1]) <= L) f[1][1][2 * (a[2] - a[1])][0] = 1;
    rep(i, 1, n) {
        int d = a[i + 2] - a[i + 1];
        rep(j, 1, i + 1) {
            rep(k, 0, L + 1) {
                rep(e, 0, min(j + 1, 3)) {
                    if(!f[i & 1][j][k][e]) continue;
                    if(e < 2) {
                        if(k + (2 * (j + 1) - (e + 1)) * d <= L)
                            add(f[(i + 1) & 1][j + 1][k + (2 * (j + 1) - (e + 1)) * d][e + 1], (2 - e) * f[i & 1][j][k][e]);
                        if(k + (2 * j - (e + 1)) * d <= L)
                            add(f[(i + 1) & 1][j][k + (2 * j - (e + 1)) * d][e + 1], (2 - e) * f[i & 1][j][k][e]);
                    }
                    if(k + (2 * (j + 1) - e) * d <= L)
                        add(f[(i + 1) & 1][j + 1][k + (2 * (j + 1) - e) * d][e], 1LL * (j - e + 1) * f[i & 1][j][k][e]);
                    if(k + (2 * j - e) * d <= L)
                        add(f[(i + 1) & 1][j][k + (2 * j - e) * d][e], 1LL * (2 * j - e) * f[i & 1][j][k][e]);
                    if(j > 1 && k + (2 * (j - 1) - e) * d <= L)
                        add(f[(i + 1) & 1][j - 1][k + (2 * (j - 1) - e) * d][e], 1LL * (j - 1) * f[i & 1][j][k][e]);
                }
            }
        }
        memset(f[i & 1], 0, sizeof f[i & 1]);
    }
    int res = 0;
    rep(i, 0, L + 1) add(res, f[n & 1][1][i][2]);
    if(n == 1) res = 1;
    cout << res << endl;
    //trau();
}

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