結果

問題 No.2283 Prohibit Three Consecutive
ユーザー miscalcmiscalc
提出日時 2023-04-28 21:37:39
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,667 bytes
コンパイル時間 4,184 ms
コンパイル使用メモリ 269,072 KB
実行使用メモリ 27,916 KB
最終ジャッジ日時 2024-11-17 20:36:17
合計ジャッジ時間 5,911 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
6,816 KB
testcase_02 AC 2 ms
6,820 KB
testcase_03 AC 262 ms
17,972 KB
testcase_04 AC 242 ms
15,124 KB
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 AC 51 ms
27,776 KB
testcase_09 AC 86 ms
27,916 KB
testcase_10 AC 47 ms
27,904 KB
testcase_11 AC 86 ms
27,792 KB
testcase_12 WA -
testcase_13 WA -
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ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using tlll = tuple<ll, ll, ll>;
constexpr ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll safemod(ll A, ll M) {ll res = A % M; if (res < 0) res += M; return res;}
ll divfloor(ll A, ll B) {if (B < 0) A = -A, B = -B; return (A - safemod(A, B)) / B;}
ll divceil(ll A, ll B) {if (B < 0) A = -A, B = -B; return divfloor(A + B - 1, B);}
ll pow_ll(ll A, ll B) {if (A == 0 || A == 1) {return A;} if (A == -1) {return B & 1 ? -1 : 1;} ll res = 1; for (int i = 0; i < B; i++) {res *= A;} return res;}
ll mul_limited(ll A, ll B, ll M = INF) { return A > M / B ? M : A * B; }
ll pow_limited(ll A, ll B, ll M = INF) { if (A == 0 || A == 1) {return A;} ll res = 1; for (int i = 0; i < B; i++) {if (res > M / A) return M; res *= A;} return res;}
ll logfloor(ll A, ll B) {assert(A >= 2); ll res = 0; for (ll tmp = 1; tmp <= B / A; tmp *= A) {res++;} return res;}
ll logceil(ll A, ll B) {assert(A >= 2); ll res = 0; for (ll tmp = 1; tmp < B; tmp *= A) {res++;} return res;}
ll arisum_ll(ll a, ll d, ll n) { return n * a + (n & 1 ? ((n - 1) >> 1) * n : (n >> 1) * (n - 1)) * d; }
ll arisum2_ll(ll a, ll l, ll n) { return n & 1 ? ((a + l) >> 1) * n : (n >> 1) * (a + l); }
ll arisum3_ll(ll a, ll l, ll d) { assert((l - a) % d == 0); return arisum2_ll(a, l, (l - a) / d + 1); }
template<class T> void unique(vector<T> &V) {V.erase(unique(V.begin(), V.end()), V.end());}
template<class T> void sortunique(vector<T> &V) {sort(V.begin(), V.end()); V.erase(unique(V.begin(), V.end()), V.end());}
#define FINALANS(A) do {cout << (A) << '\n'; exit(0);} while (false)
template<class T> void printvec(const vector<T> &V) {int _n = V.size(); for (int i = 0; i < _n; i++) cout << V[i] << (i == _n - 1 ? "" : " ");cout << '\n';}
template<class T> void printvect(const vector<T> &V) {for (auto v : V) cout << v << '\n';}
template<class T> void printvec2(const vector<vector<T>> &V) {for (auto &v : V) printvec(v);}
//*
#include <atcoder/all>
using namespace atcoder;
using mint = modint998244353;
//using mint = modint1000000007;
//using mint = modint;
//*/

bool sub(ll N, string S)
{
  vector dp(N + 1, vector(3, vector(3, false)));
  dp.at(0).at(0).at(2) = true;
  for (ll i = 0; i < N; i++)
  {
    for (ll j = 0; j < 3; j++)
    {
      for (ll k = 0; k < 3; k++)
      {
        if (!dp.at(i).at(j).at(k))
          continue;
        for (ll nk = 0; nk < 2; nk++)
        {
          if (nk == 0 && S.at(i) == '1')
            continue;
          if (nk == 1 && S.at(i) == '0')
            continue;

          ll nj = (k == nk ? j + 1 : 1);
          if (nj == 3)
            continue;
          dp.at(i + 1).at(nj).at(nk) = true;
        }
      }
    }
  }

  for (ll j = 0; j < 3; j++)
  {
    for (ll k = 0; k < 2; k++)
    {
      if (dp.at(N).at(j).at(k))
        return true;
    }
  }
  return false;
}

bool solve(ll N, string S)
{
  for (ll s1 = 0; s1 < 2; s1++)
  {
    for (ll s2 = 0; s2 < 2; s2++)
    {
      string T = S;
      if (T.at(0) == '?')
        T.at(0) = '0' + s1;
      if (T.at(1) == '?')
        T.at(1) = '0' + s2;
      if (T.at(N - 2) == T.at(N - 1) && T.at(N - 1) == T.at(0))
        continue;
      if (T.at(N - 1) == T.at(0) && T.at(0) == T.at(1))
        continue;
      if (sub(N, T))
        return true;
    }
  }
  return false;
}

int main()
{
  ll T;
  cin >> T;
  while (T--)
  {
    ll N;
    string S;
    cin >> N >> S;
    cout << (solve(N, S) ? "Yes" : "No") << endl;
  }
}
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