Submission #902576

#	Time	Username	Problem	Language	Result	Execution time	Memory
902576		Spade1	Skyscraper (JOI16_skyscraper)	C++14	20 / 100	708 ms	495116 KiB

skyscraper

#include <bits/stdc++.h>
#define ll long long
#define pii pair<int, int>
#define pll pair<ll ll>
#define st first
#define nd second
#define pb push_back
using namespace std;

const int maxN = 1e2 + 10;
const int maxL = 1e3 + 10;
const int M = 1e9 + 7;

ll dp[maxN][maxN][4*maxL][3], a[maxN];

void solve() {
    ll n, l; cin >> n >> l;
    if (n == 1) {
        cout << 1 << '\n';
        return;
    }
    for (int i = 1; i <= n; ++i) cin >> a[i];
    sort(a+1, a+n+1);
    dp[0][0][2*l][0] = 1;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            for (int k = 0; k <= 4*l; ++k) {
                if (k+2*a[i] <= 4*l) {
                    dp[i][j][k][0] = (dp[i][j][k][0] + j*dp[i-1][j-1][k+2*a[i]][0]) % M;
                    dp[i][j][k][1] = (dp[i][j][k][1] + (j-1)*dp[i-1][j-1][k+2*a[i]][1]) % M;
                    dp[i][j][k][2] = (dp[i][j][k][2] + (j-2)*dp[i-1][j-1][k+2*a[i]][2]) % M;
                }
                if (k+a[i] <= 4*l) {
                    dp[i][j][k][1] = (dp[i][j][k][1] + 2*dp[i-1][j-1][k+a[i]][0]) % M;
                    dp[i][j][k][2] = (dp[i][j][k][2] + dp[i-1][j-1][k+a[i]][1]) % M;
                }
                dp[i][j][k][0] = (dp[i][j][k][0] + (2*j)*dp[i-1][j][k][0]) % M;
                dp[i][j][k][1] = (dp[i][j][k][1] + (2*j-1)*dp[i-1][j][k][1]) % M;
                dp[i][j][k][2] = (dp[i][j][k][2] + (2*j-2)*dp[i-1][j][k][2]) % M;
                if (k-a[i] >= 0) {
                    dp[i][j][k][1] = (dp[i][j][k][1] + 2*dp[i-1][j][k-a[i]][0]) % M;
                    dp[i][j][k][2] = (dp[i][j][k][2] + dp[i-1][j][k-a[i]][1]) % M;
                }
                if (k-2*a[i] >= 0) {
                    dp[i][j][k][0] = (dp[i][j][k][0] + j*dp[i-1][j+1][k-2*a[i]][0]) % M;
                    dp[i][j][k][1] = (dp[i][j][k][1] + j*dp[i-1][j+1][k-2*a[i]][1]) % M;
                    dp[i][j][k][2] = (dp[i][j][k][2] + j*dp[i-1][j+1][k-2*a[i]][2]) % M;
                }
            }
        }
    }
    ll ans = 0;
    for (int i = 2*l; i <= 3*l; ++i) {
        ans += dp[n][1][i][2];
        ans %= M;
    }
    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(NULL);
    int t = 1;
//    cin >> t;
    while (t--) solve();
}

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3: 0 / 80