Submission #259290

#	Time	Username	Problem	Language	Result	Execution time	Memory
259290		Bruteforceman	Ruins 3 (JOI20_ruins3)	C++11	58 / 100	6082 ms	23804 KiB

ruins3

#include <bits/stdc++.h>
using namespace std;
const int maxn = 605 * 2;
const long long mod = 1e9 + 7;
long long ways[maxn];
int a[maxn];

long long binpow(long long x, long long y) {
  long long ans = 1;
  long long power = x;
  while(y) {
    if(y & 1) ans = (ans * power) % mod;
    power = (power * power) % mod;
    y >>= 1;
  }
  return ans;
}
long long inv(long long x) {
  return binpow(x % mod, mod - 2);
}
int var;
long long mem[maxn][maxn];
long long C[maxn][maxn];
long long fac[] = {1, 1, 2};

long long calc(int cur, int take) {
  long long ans = 0;
  if(cur == 1) {
    for(int i = 0; i <= 2; i++) {
      if(take + i == var + 1) {
        return (C[take + i][i] * inv(fac[2 - i])) % mod;
      }
    }
    return 0;
  }
  if(mem[cur][take] != -1) return mem[cur][take];
  for(int i = 0; i <= 2; i++) {
    if(take + i <= var - cur + 1) {
      long long add = (calc(cur - 1, take + i) * C[take + i][i]) % mod;
      ans += (add * inv(fac[2 - i])) % mod;
      ans %= mod;
    }
  }
  return mem[cur][take] = ans;
}
int n;
long long fn[maxn][maxn];
long long dp(int cur, int bound) {
  if(cur == n) return (bound == 0);
  if(fn[cur][bound] != -1) return fn[cur][bound];
  long long ans = (dp(cur + 1, bound) * (n - bound - cur)) % mod;
  for(int i = 0; i < bound; i++) {
    int free = a[cur + 1] - (n - bound + (cur + 1));
    if(free < bound - i) continue;
    ans += ((C[free][bound - i] * 
      ways[bound - i]) % mod) * dp(cur + 1, i);
    ans %= mod;
  }
  return fn[cur][bound] = ans;
}

int main() {
  cin >> n;
  for(int i = 1; i <= n; i++) cin >> a[i];
  for(int i = 0; i <= n + 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]) % mod;
    }
  }
  for(int i = 1; i <= n; i++) {
    memset(mem, -1, sizeof mem);
    var = i;
    ways[i] = calc(var, 0);
  }
  memset(fn, -1, sizeof fn);
  cout << ((a[n] != n + n) ? 0 : dp(0, n)) << endl;
  return 0;
}

• Subtask 1: 6 / 6
• Subtask 2: 52 / 52
• Subtask 3: 0 / 42