Submission #212789

#	Time	Username	Problem	Language	Result	Execution time	Memory
212789		IOrtroiii	Ruins 3 (JOI20_ruins3)	C++14	100 / 100	202 ms	1920 KiB

ruins3

#include <bits/stdc++.h>

using namespace std;

const int MOD = 1e9 + 7;
using ll = long long;

int mult(int x, int y) {
   return (long long) x * y % MOD;
}

void add(int &x, int y) {
   x += y;
   if (x >= MOD) x -= MOD;
}

int main() {
   ios_base::sync_with_stdio(false);
   int N;
   cin >> N;
   vector<bool> alive(2 * N);
   for (int i = 0; i < N; ++i) {
      int x;
      cin >> x;
      alive[--x] = true;
   }
   vector<int> ways(N + 1);
   {
      vector<vector<int>> dp(N + 1, vector<int>(N + 1));
      dp[0][0] = 1;
      for (int i = 1; i <= N; ++i) {
         for (int j = 0; j <= i; ++j) {
            dp[i][j] = dp[i - 1][j];
            if (j > 0) add(dp[i][j], mult(2 * j, dp[i - 1][j - 1]));
            if (j > 1) add(dp[i][j], mult(j * (j - 1), dp[i - 1][j - 2]));
         }
      }
      for (int i = 1; i <= N; ++i) {
         ways[i] = mult(dp[i - 1][i - 1], i + 1);
      }
   }
   vector<vector<int>> C(N + 1, vector<int>(N + 1));
   for (int i = 0; i <= N; ++i) {
      C[i][0] = 1;
      for (int j = 1; j <= i; ++j) {
         C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
         if (C[i][j] >= MOD) C[i][j] -= MOD;
      }
   }
   vector<int> dp(N + 1);
   dp[0] = 1;
   int cnt0 = 0;
   int cnt1 = 0;
   for (int i = 2 * N - 1; i >= 0; --i) {
      vector<int> ndp(N + 1);
      if (alive[i]) {
         for (int j = cnt0; j <= cnt1; ++j) {
            if (!dp[j]) continue;
            add(ndp[j], dp[j]);
            for (int k = j + 1; k <= cnt1 + 1; ++k) {
               add(ndp[k], mult(mult(C[cnt1 - j][k - j - 1], ways[k - j]), dp[j]));
            }
         }
         ++cnt1;
      } else {
         for (int j = cnt0; j <= cnt1; ++j) {
            add(ndp[j], mult(j - cnt0, dp[j]));
         }
         ++cnt0;
      }
      dp = ndp;
   }
   int ans = dp[N];
   for (int i = 0; i < N; ++i) ans = mult(ans, (MOD + 1) >> 1);
   cout << ans << "\n";
   return 0;
}

• Subtask 1: 6 / 6
• Subtask 2: 52 / 52
• Subtask 3: 42 / 42