結果

問題 No.474 色塗り２
ユーザー fumiphysfumiphys
提出日時 2019-08-15 21:00:40
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 217 ms / 2,000 ms
コード長 4,301 bytes
コンパイル時間 1,838 ms
コンパイル使用メモリ 168,540 KB
実行使用メモリ 38,796 KB
最終ジャッジ日時 2023-10-19 19:24:35
合計ジャッジ時間 2,790 ms
ジャッジサーバーID
（参考情報） 		judge12 / judge13
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 23 ms
38,736 KB
testcase_01 AC 23 ms
38,792 KB
testcase_02 AC 23 ms
38,796 KB
testcase_03 AC 217 ms
38,796 KB
testcase_04 AC 23 ms
38,792 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 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 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;

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;}

//  types
using ll = long long int;
using P = pair<int, int>;
using Pli = pair<ll, int>;
using Pil = pair<int, ll>;
using Pll = pair<ll, ll>;
using Pdd = pair<double, double>;
using cd = complex<double>;

// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = (1 << 20);
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
template<typename T>
T extgcd(T a, T b, T &x, T &y){ 
  T d = a;
  if(b != 0){
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  }else{
    x = 1, y = 0;
  }
  return d;
}

template <typename T>
T modinv(T a, T m){
  long long x = 0, y = 0;
  extgcd<long long>(a, m, x, y);
  x %= m;
  if(x < 0)x += m;
  return x;
}

ll fac[3000100];
int T[3000100];

ll comb(ll n, ll k){
  int t = T[n] - T[k] - T[n-k];
  ll res = (t < 20 ? (1 << t): 0);
  res = res * fac[n] % mod;
  res = res * modinv(fac[k], (ll)mod) % mod;
  res = res * modinv(fac[n-k], (ll)mod) % mod;
  return res;
}

void init(){
  fac[0] = 1;
  T[0] = 0;
  for(int i = 1; i <= 3000000; i++){
    T[i] = T[i-1];
    int tmp = i;
    while(tmp % 2 == 0){
      tmp /= 2;
      T[i]++;
    }
    fac[i] = fac[i-1] * tmp % mod;
  }
}

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  init();
  INT(t);
  rep(i, t){
    INT(a); INT(b); INT(c);
    if(c % 2 == 0){
      cout << 0 << endl;
      continue;
    }
    ll A = (a + c * comb(b + c - 1, c - 1) % mod - 1) % mod;
    if((A & a) == a)cout << 1 << endl;
    else cout << 0 << endl;
  }
  return 0;
}
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