結果

問題 No.695 square1001 and Permutation 4
コンテスト
ユーザー fumiphys
提出日時 2019-07-02 22:49:58
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,583 ms / 7,000 ms
コード長 4,569 bytes
コンパイル時間 1,748 ms
コンパイル使用メモリ 173,836 KB
実行使用メモリ 42,200 KB
最終ジャッジ日時 2024-07-22 19:37:25
合計ジャッジ時間 11,676 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 12
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 ll mod = 100000000000000007LL;
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;}

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>
pair<T, T> chinese_reminder_theorem(vector<T> b, vector<T> m){
  T r = 0, M = 1;
  for(int i = 0; i < b.size(); i++){
    T x, y;
    T d = extgcd<T>(M, m[i], x, y);
    if((b[i] - r) % d != 0)return make_pair(0, -1);
    T tmp = (b[i] - r) / d * x % (m[i] / d);
    r += M * tmp;
    M *= m[i] / d;
  }
  r %= M;
  if(r < 0)r += M;
  return make_pair(r % M, M);
}

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  LL(n); INT(m); vector<ll> x(m); cin >> x;
  vector<ll> a(2); a[0] = 168647939LL; a[1] = 592951213LL;
  vector<ll> res(2, 0);
  rep(i, 2){
    vector<int> dp((n+1)/2 + 1, 0);
    dp[1] = 1;
    FOR(j, 2, sz(dp)){
      rep(k, m){
        if(j - x[k] > 0)dp[j] = (dp[j] + dp[j - x[k]]) % a[i];
      }
    }
    for(int j = 1; j < sz(dp); j++){
      rep(k, m){
        if(j + x[k] >= sz(dp) && j + x[k] <= n){
          res[i] = (res[i] + (ll)dp[j] * (ll)dp[n - j - x[k] + 1]) % a[i];
        }
      }
    }
  }
  auto p = chinese_reminder_theorem<ll>(res, a);
  cout << p.FI << endl;
  return 0;
}
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