結果

問題 No.663 セルオートマトンの逆操作
ユーザー fumiphysfumiphys
提出日時 2019-07-02 19:53:35
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 4,269 bytes
コンパイル時間 1,728 ms
コンパイル使用メモリ 172,492 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-09-13 13:58:01
合計ジャッジ時間 2,667 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
6,816 KB
testcase_01 AC 2 ms
6,944 KB
testcase_02 AC 2 ms
6,940 KB
testcase_03 AC 2 ms
6,944 KB
testcase_04 AC 2 ms
6,940 KB
testcase_05 AC 2 ms
6,940 KB
testcase_06 AC 2 ms
6,940 KB
testcase_07 AC 2 ms
6,944 KB
testcase_08 AC 2 ms
6,944 KB
testcase_09 AC 2 ms
6,944 KB
testcase_10 AC 2 ms
6,940 KB
testcase_11 AC 2 ms
6,944 KB
testcase_12 AC 2 ms
6,944 KB
testcase_13 AC 2 ms
6,940 KB
testcase_14 AC 2 ms
6,940 KB
testcase_15 AC 2 ms
6,940 KB
testcase_16 AC 2 ms
6,940 KB
testcase_17 AC 2 ms
6,944 KB
testcase_18 AC 2 ms
6,940 KB
testcase_19 AC 2 ms
6,944 KB
testcase_20 AC 2 ms
6,940 KB
testcase_21 AC 2 ms
6,940 KB
testcase_22 AC 2 ms
6,940 KB
testcase_23 AC 3 ms
6,940 KB
testcase_24 AC 2 ms
6,940 KB
testcase_25 AC 2 ms
6,944 KB
testcase_26 AC 3 ms
6,940 KB
testcase_27 AC 2 ms
6,940 KB
testcase_28 AC 2 ms
6,940 KB
権限があれば一括ダウンロードができます

ソースコード

diff # 

// includes
#include <bits/stdc++.h>

// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define vrep(v, i) for(int i = 0; i < (v).size(); i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define FI first
#define SE second
#define dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n];
#define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m];
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;

//  types
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef complex<double> cd;

// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}

ll dp[2020][4][4];

int calc(int l, int r, int C){
  if((l & 1) != (r >> 1))return 0;
  int a = (l >> 1), b = l & 1, c = r & 1;
  if(b == 0 && c == 0)return 1 - C;
  if(a == 1 && b == 1 && c == 1)return 1 - C;
  return C;
}

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n); vector<int> c(n); cin >> c;

  rep(i, 4)rep(j, 4)dp[2][j][j] = 1;
  FOR(i, 2, n){
    rep(j, 4){
      rep(k, 4){
        if(i + 1 == n){
          rep(l, 4){
            dp[i+1][j][k] = (dp[i+1][j][k] + dp[i][j][l] * calc(l, k, c[i-1]) * calc(k, ((k&1) << 1) + (j >> 1), c[n-1]) * calc(((k&1) << 1) + (j >> 1), j, c[0])) % mod;
          }
        }else{
          rep(l, 4){
            dp[i+1][j][k] = (dp[i+1][j][k] + dp[i][j][l] * calc(l, k, c[i-1])) % mod;
          }
        }
      }
    }
  }

  ll res = 0;
  rep(i, 4)rep(j, 4){
    res = (res + dp[n][i][j]) % mod;
  }
  if(res < 0)res += mod;
  cout << res << endl;
  return 0;
}
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