結果
| 問題 |
No.42 貯金箱の溜息
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2020-04-11 00:02:45 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 42 ms / 5,000 ms |
| コード長 | 6,384 bytes |
| コンパイル時間 | 1,866 ms |
| コンパイル使用メモリ | 176,108 KB |
| 実行使用メモリ | 5,376 KB |
| 最終ジャッジ日時 | 2024-09-16 03:54:30 |
| 合計ジャッジ時間 | 2,220 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 3 |
ソースコード
// includes
#include <bits/stdc++.h>
using namespace std;
// macros
#define pb emplace_back
#define mk make_pair
#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 whole(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bit(n) (1LL<<(n))
// functions
template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}
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>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000009;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
fast_io(){ios_base::sync_with_stdio(false); cin.tie(nullptr); cout << fixed << setprecision(20);}
} fast_io_;
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;
}
template <typename T>
T power(T a, T n, T mod) {
T res = 1;
T tmp = n;
T curr = a;
while(tmp){
if(tmp % 2 == 1){
res = (T)(res * curr % mod);
}
curr = (T)(curr * curr % mod);
tmp >>= 1;
}
return res;
}
struct Mint{
const static ll default_mod = (ll)(1e9 + 9);
ll MOD = default_mod;
ll x = 0;
ll get_mod(){
return MOD;
}
Mint(){}
Mint(ll x_, ll MOD=default_mod): MOD(MOD){
x = x_;
x %= MOD;
if(x < 0)x += MOD;
}
Mint(const Mint &m){
x = m.x;
MOD = m.MOD;
}
Mint &operator+=(const Mint &y){
x = (x + y.x) % MOD;
if(x < 0)x += MOD;
return *this;
}
Mint &operator-=(const Mint &y){
x = (x - y.x) % MOD;
if(x < 0)x += MOD;
return *this;
}
Mint &operator*=(const Mint &y){
x = (x * y.x) % MOD;
if(x < 0)x += MOD;
return *this;
}
Mint inverse() const{
return Mint(modinv<ll>(x, MOD), MOD);
}
Mint &operator/=(const Mint &y){
x = (x * y.inverse().x) % MOD;
if(x < 0)x += MOD;
return *this;
}
Mint operator-() const{
return Mint(-x, MOD);
}
Mint operator+(const Mint &y) const{
return Mint(*this) += y;
}
Mint operator-(const Mint &y) const{
return Mint(*this) -= y;
}
Mint operator*(const Mint &y) const{
return Mint(*this) *= y;
}
Mint operator/(const Mint &y) const{
return Mint(*this) /= y;
}
bool operator==(const Mint &y) const{
return x == y.x;
}
bool operator!=(const Mint &y) const{
return x != y.x;
}
Mint pow(long long k) const{
long long ret = power<long long>(x, k, MOD);
return Mint(ret, MOD);
}
friend ostream& operator<<(ostream &os, const Mint &m){
return os << m.x;
}
friend istream& operator>>(istream &is, Mint &m){
ll t;
is >> t;
m = Mint(t);
return is;
}
explicit operator long long() const{
return x;
}
};
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;
}
};
int d[6] = {1, 5, 10, 50, 100, 500};
ll dp[10010];
int main(int argc, char const* argv[])
{
dp[0] = 1;
rep(i, 6){
rep(j, 10010){
if(j-d[i]>=0)(dp[j] += dp[j-d[i]]) %= mod;
}
}
vector<LagrangeInterpolationM<Mint>> rip(500);
rep(i, 500){
vector<Mint> x(7), y(7);
rep(j, 7){
x[j] = j;
y[j] = dp[i + 500 * j];
}
rip[i] = LagrangeInterpolationM<Mint>(x, y);
}
int t; cin >> t;
rep(i_, t){
ll m; cin >> m;
cout << rip[m%500].val(m/500) << endl;
}
return 0;
}