結果

問題 No.129 お年玉(2)
ユーザー fumiphysfumiphys
提出日時 2019-05-23 17:16:56
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 96 ms / 5,000 ms
コード長 3,707 bytes
コンパイル時間 1,830 ms
コンパイル使用メモリ 166,460 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-05-06 00:02:58
合計ジャッジ時間 4,826 ms
ジャッジサーバーID
（参考情報） 		judge2 / 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 6 ms
5,248 KB
testcase_01 AC 91 ms
5,376 KB
testcase_02 AC 1 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 AC 2 ms
5,376 KB
testcase_05 AC 38 ms
5,376 KB
testcase_06 AC 60 ms
5,376 KB
testcase_07 AC 74 ms
5,376 KB
testcase_08 AC 49 ms
5,376 KB
testcase_09 AC 72 ms
5,376 KB
testcase_10 AC 79 ms
5,376 KB
testcase_11 AC 50 ms
5,376 KB
testcase_12 AC 65 ms
5,376 KB
testcase_13 AC 68 ms
5,376 KB
testcase_14 AC 64 ms
5,376 KB
testcase_15 AC 52 ms
5,376 KB
testcase_16 AC 59 ms
5,376 KB
testcase_17 AC 71 ms
5,376 KB
testcase_18 AC 66 ms
5,376 KB
testcase_19 AC 72 ms
5,376 KB
testcase_20 AC 71 ms
5,376 KB
testcase_21 AC 90 ms
5,376 KB
testcase_22 AC 50 ms
5,376 KB
testcase_23 AC 78 ms
5,376 KB
testcase_24 AC 76 ms
5,376 KB
testcase_25 AC 48 ms
5,376 KB
testcase_26 AC 51 ms
5,376 KB
testcase_27 AC 48 ms
5,376 KB
testcase_28 AC 89 ms
5,376 KB
testcase_29 AC 2 ms
5,376 KB
testcase_30 AC 2 ms
5,376 KB
testcase_31 AC 2 ms
5,376 KB
testcase_32 AC 2 ms
5,376 KB
testcase_33 AC 2 ms
5,376 KB
testcase_34 AC 1 ms
5,376 KB
testcase_35 AC 1 ms
5,376 KB
testcase_36 AC 2 ms
5,376 KB
testcase_37 AC 2 ms
5,376 KB
testcase_38 AC 11 ms
5,376 KB
testcase_39 AC 16 ms
5,376 KB
testcase_40 AC 46 ms
5,376 KB
testcase_41 AC 28 ms
5,376 KB
testcase_42 AC 13 ms
5,376 KB
testcase_43 AC 13 ms
5,376 KB
testcase_44 AC 96 ms
5,376 KB
testcase_45 AC 8 ms
5,376 KB
testcase_46 AC 19 ms
5,376 KB
testcase_47 AC 89 ms
5,376 KB
testcase_48 AC 54 ms
5,376 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 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 int mod = 1e9;
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;}

ll dp[2][10001];

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  LL(n); LL(m);
  int rem = n / 1000 - ((n / 1000) / m) * m;
  dp[0][0] = 1;
  rep(i, m+1){
    rep(j, i + 1){
      if(j>0 && j<i)dp[(i+1)&1][j] = (dp[i&1][j] + dp[i&1][j-1]) % mod;
      else dp[(i+1)&1][j] = 1;
      //cerr << i << " " << j << " " << dp[(i+1)&1][j] << endl;
    }
  }
  cout << dp[(m+1)&1][rem] << endl;
  return 0;
}
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