Submission #659768

#	Time	Username	Problem	Language	Result	Execution time	Memory
659768		MohammadAghil	Fences (JOI18_fences)	C++17	0 / 100	27 ms	46680 KiB

fences

      #include <bits/stdc++.h>
//   #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
 using namespace std;
  typedef long long ll;
   typedef pair<int, int> pp;
     #define per(i,r,l) for(int i = (r); i >= (l); i--)
       #define rep(i,l,r) for(int i = (l); i < (r); i++)
          #define all(x) begin(x), end(x)
             #define sz(x) (int)(x).size()
                 #define pb push_back
                     #define ss second
                          #define ff first
                                  void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
void IOS(){
     cin.tie(0) -> sync_with_stdio(0);
     // #ifndef ONLINE_JUDGE
     //      #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
     //      freopen("inp.txt", "r", stdin);
     //      freopen("out.txt", "w", stdout);
     // #else
     //      #define er(args ...) 0
     // #endif
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 3e3 + 5, lg = 22, inf = ll(1e9) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll,md);return k*k%md*(b&1ll?a:1)%md;}

ll f[maxn][maxn];
ll C[maxn][maxn], G[maxn];
 
int main(){ IOS();
     int n, m; cin >> n >> m;
     rep(i,0,n+1){
          rep(j,0,m+1){
               f[i][j] = 1;
               if(j && i) f[i][j] = (f[i][j-1] + 4ll*i*f[i-1][j-1])%mod;
          }
     }
     rep(i,0,maxn){
          C[i][0] = 1;
          rep(j,1,i+1) C[i][j] = (C[i-1][j] + C[i-1][j-1])%mod;
     }
     G[0] = 1;
     rep(i,1,maxn){
          G[i] = 1ll*G[i-1]*i%mod*(2*i-1)%mod;
     }
     ll ans = mod - 1;
     rep(a,0,n+1){
          rep(b,0,m+1){
               if(2*a + b <= n && 2*b + a <= m){
                    ans += f[n - 2*a - b][m - 2*b - a]*C[m][a]%mod*C[n][2*a]%mod*G[a]%mod*C[n-2*a][b]%mod*C[m-a][2*b]%mod*G[b]%mod;
                    ans %= mod;
               }
          }
     }
     cout << ans << '\n';
     return 0-0;
}

• Subtask 1: 0 / 18
• Subtask 2: 0 / 33
• Subtask 3: 0 / 49