結果

問題 No.42 貯金箱の溜息
ユーザー fumiphysfumiphys
提出日時 2019-08-21 23:10:58
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 28 ms / 5,000 ms
コード長 6,011 bytes
コンパイル時間 1,787 ms
コンパイル使用メモリ 176,912 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-10-10 04:04:55
合計ジャッジ時間 2,507 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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 24 ms
6,820 KB
testcase_01 AC 28 ms
6,816 KB
testcase_02 AC 27 ms
6,816 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 = 1e9 + 9;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
template <int MOD=int(1e9+9)>
struct ModInt {
  ll x = 0;
  ModInt(){}
  ModInt(ll x_){
    x = int(x_ % MOD);
    if(x < 0)x += MOD;
  }
  ModInt(const ModInt &m){
    x = m.x;
  }
  ModInt& operator+=(const ModInt &y){
    x = (x + y.x) % MOD;
    if(x < 0)x += MOD;
    return *this;
  }
  ModInt& operator-=(const ModInt &y){
    x = (x - y.x) % MOD;
    if(x < 0)x += MOD;
    return *this;
  }
  ModInt& operator*=(const ModInt &y){
    x = (x * y.x) % MOD;
    if(x < 0)x += MOD;
    return *this;
  }
  ModInt& operator/=(const ModInt &y){
    x = (x * y.inverse().x) % MOD;
    if(x < 0)x += MOD;
    return *this;
  }
  ModInt inverse() const{
    ll res = 1;
    ll tmp = MOD - 2;
    ll curr = x;
    while(tmp){
      if(tmp % 2 == 1)res = res * curr % MOD;
      curr = curr * curr % MOD;
      tmp /= 2;
    }
    return ModInt(res);
  }
  ModInt operator-() const{
    return ModInt(-x);
  }
  ModInt operator+(const ModInt &y) const{
    return ModInt(*this) += y;
  }
  ModInt operator-(const ModInt &y) const{
    return ModInt(*this) -= y;
  }
  ModInt operator*(const ModInt &y) const{
    return ModInt(*this) *= y;
  }
  ModInt operator/(const ModInt &y) const{
    return ModInt(*this) /= y;
  }
  bool operator==(const ModInt &y) const{
    return x == y.x;
  }
  bool operator!=(const ModInt &y) const{
    return x != y.x;
  }
  friend ostream& operator<<(ostream &os, const ModInt<MOD> &m){
    return os << m.x;
  }
  friend istream& operator>>(istream &is, ModInt<MOD> &m){
    long t;
    is >> t;
    m = ModInt<MOD>(t);
    return is;
  }
};

template <typename T>
struct LagrangeInterpolationM{
  int n = 0;
  vector<T> x, y;
  vector<T> nume;
  LagrangeInterpolationM(){}
  LagrangeInterpolationM(const vector<T> &x, const vector<T> &y): x(x), y(y){
    n = x.size() - 1;
    nume.resize(n + 1);
    for(int i = 0; i <= n; i++){
      T t = T(1);
      for(int j = 0; j <= n; j++){
        if(i == j)continue;
        t = t * (x[i] - x[j]);
      }
      nume[i] = t.inverse();
    }
  }
  T val(T t){
    T a = T(1);
    for(int i = 0; i <= n; i++){
      if(t == x[i])return y[i];
      a = a * (t - x[i]);
    }
    T res = T(0);
    for(int i = 0; i <= n; i++){
      res += y[i] * nume[i] * (a / (t - x[i]));
    }
    return res;
  }
};
ll dp[3100];
int a[6] = {1, 5, 10, 50, 100, 500};


int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  dp[0] = 1;
  rep(i, 6){
    rep(j, 3001){
      dp[j] = (dp[j] + (j-a[i]>=0?dp[j-a[i]]: 0)) % mod;
    }
  }
  vector<LagrangeInterpolationM<ModInt<>>> lim(500);
  rep(i, 500){
    vector<ModInt<>> x, y;
    rep(j, 6){
      x.pb(j);
      y.pb(dp[i+j*500]);
    }
    lim[i] = LagrangeInterpolationM<ModInt<>>(x, y);
  }
  INT(t);
  rep(i_, t){
    LL(m);
    ll l = m / 500;
    ll mo = m % 500;
    cout << lim[mo].val(l) << endl;
  }
  return 0;
}
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