Submission #328740

#	Time	Username	Problem	Language	Result	Execution time	Memory
328740		EndRay	Skyscraper (JOI16_skyscraper)	C++17	100 / 100	285 ms	128128 KiB

skyscraper

#include<bits/stdc++.h>
#define ll long long

using namespace std;

const int N = 100+1, L = 1000+1, mod = 1e9+7;

int n, l;
int a[N];
ll dp[N][N][L][3];

int main(){
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    cin >> n >> l;
    for(int i = 0; i < n; ++i)
        cin >> a[i];
    if(n == 1){
        cout << "1\n";
        return 0;
    }
    sort(a, a+n);
    a[n] = L;
    dp[0][0][0][0] = 1;
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= n; ++j){
            for(int k = 0; k <= l; ++k){
                int dif;
                // e = 0
                dif = 2*j*(a[i]-a[i-1]);
                if(dif <= k){
                    dp[i][j][k][0] = (dp[i][j][k][0] + dp[i-1][j-1][k - dif][0]*j)%mod;
                    dp[i][j][k][0] = (dp[i][j][k][0] + dp[i-1][j][k - dif][0]*2*j)%mod;
                    dp[i][j][k][0] = (dp[i][j][k][0] + dp[i-1][j+1][k - dif][0]*j)%mod;
                }
                // e = 1
                dif = (2*j - 1)*(a[i] - a[i-1]);
                if(dif <= k){
                    dp[i][j][k][1] = (dp[i][j][k][1] + dp[i-1][j-1][k - dif][0]*2)%mod;
                    dp[i][j][k][1] = (dp[i][j][k][1] + dp[i-1][j][k - dif][0]*2)%mod;
                    dp[i][j][k][1] = (dp[i][j][k][1] + dp[i-1][j-1][k - dif][1]*(j-1))%mod;
                    dp[i][j][k][1] = (dp[i][j][k][1] + dp[i-1][j][k - dif][1]*(2*j-1))%mod;
                    dp[i][j][k][1] = (dp[i][j][k][1] + dp[i-1][j+1][k - dif][1]*j)%mod;
                }
                // e = 2
                dif = (2*j - 2)*(a[i] - a[i-1]);
                if(dif <= k){
                    dp[i][j][k][2] = (dp[i][j][k][2] + dp[i-1][j-1][k - dif][1])%mod;
                    dp[i][j][k][2] = (dp[i][j][k][2] + dp[i-1][j][k - dif][1])%mod;
                    dp[i][j][k][2] = (dp[i][j][k][2] + dp[i-1][j-1][k - dif][2]*(j-2))%mod;
                    dp[i][j][k][2] = (dp[i][j][k][2] + dp[i-1][j][k - dif][2]*(2*j-2))%mod;
                    dp[i][j][k][2] = (dp[i][j][k][2] + dp[i-1][j+1][k - dif][2]*j)%mod;
                }
            }
        }
    }
    ll ans = 0;
    for(int i = 0; i <= l; ++i)
        ans = (ans + dp[n][1][i][2])%mod;
    cout << ans << "\n";
}

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