結果

問題 No.269 見栄っ張りの募金活動
ユーザー c_elegans_ayanac_elegans_ayana
提出日時 2018-11-14 21:09:37
言語 C++11
(gcc 11.4.0)
結果
AC  
実行時間 19 ms / 5,000 ms
コード長 5,498 bytes
コンパイル時間 1,893 ms
コンパイル使用メモリ 158,924 KB
実行使用メモリ 11,144 KB
最終ジャッジ日時 2023-09-23 01:36:34
合計ジャッジ時間 2,682 ms
ジャッジサーバーID
(参考情報)
judge14 / judge12
このコードへのチャレンジ(β)

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 1 ms
4,380 KB
testcase_01 AC 2 ms
4,380 KB
testcase_02 AC 2 ms
4,380 KB
testcase_03 AC 3 ms
4,380 KB
testcase_04 AC 9 ms
6,284 KB
testcase_05 AC 1 ms
4,380 KB
testcase_06 AC 2 ms
4,380 KB
testcase_07 AC 19 ms
11,144 KB
testcase_08 AC 2 ms
4,380 KB
testcase_09 AC 10 ms
7,124 KB
testcase_10 AC 2 ms
4,376 KB
testcase_11 AC 4 ms
4,376 KB
testcase_12 AC 8 ms
6,316 KB
testcase_13 AC 4 ms
4,396 KB
testcase_14 AC 2 ms
4,376 KB
testcase_15 AC 4 ms
4,376 KB
testcase_16 AC 8 ms
6,068 KB
testcase_17 AC 3 ms
4,380 KB
testcase_18 AC 12 ms
8,740 KB
testcase_19 AC 5 ms
4,924 KB
testcase_20 AC 2 ms
4,380 KB
testcase_21 AC 2 ms
4,380 KB
testcase_22 AC 3 ms
4,380 KB
testcase_23 AC 2 ms
4,376 KB
testcase_24 AC 2 ms
4,376 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 #

#include <bits/stdc++.h>
#include <sys/time.h>
using namespace std;

// hamko utils
#define rep(i,n) for(long long i = 0; i < (long long)(n); i++)
#define repi(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mt make_tuple
#define mp make_pair
template<class T1, class T2> bool chmin(T1 &a, T2 b) { return b < a && (a = b, true); }
template<class T1, class T2> bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }

using ll = long long; using vll = vector<ll>; using vvll = vector<vll>; using P = pair<ll, ll>;
using ld = long double;  using vld = vector<ld>; 
using vi = vector<int>; using vvi = vector<vi>; vll conv(vi& v) { vll r(v.size()); rep(i, v.size()) r[i] = v[i]; return r; }

inline void input(int &v){ v=0;char c=0;int p=1; while(c<'0' || c>'9'){if(c=='-')p=-1;c=getchar();} while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();} v*=p; } // これを使うならば、tieとかを消して!!
template <typename T, typename U> ostream &operator<<(ostream &o, const pair<T, U> &v) {  o << "(" << v.first << ", " << v.second << ")"; return o; }
template<size_t...> struct seq{}; template<size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...>{}; template<size_t... Is> struct gen_seq<0, Is...> : seq<Is...>{};
template<class Ch, class Tr, class Tuple, size_t... Is>
void print_tuple(basic_ostream<Ch,Tr>& os, Tuple const& t, seq<Is...>){ using s = int[]; (void)s{0, (void(os << (Is == 0? "" : ", ") << get<Is>(t)), 0)...}; }
template<class Ch, class Tr, class... Args> 
auto operator<<(basic_ostream<Ch, Tr>& os, tuple<Args...> const& t) -> basic_ostream<Ch, Tr>& { os << "("; print_tuple(os, t, gen_seq<sizeof...(Args)>()); return os << ")"; }
ostream &operator<<(ostream &o, const vvll &v) { rep(i, v.size()) { rep(j, v[i].size()) o << v[i][j] << " "; o << endl; } return o; }
template <typename T> ostream &operator<<(ostream &o, const vector<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]";  return o; }
template <typename T> ostream &operator<<(ostream &o, const deque<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const unordered_set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U>  ostream &operator<<(ostream &o, const map<T, U> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U, typename V>  ostream &operator<<(ostream &o, const unordered_map<T, U, V> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it; o << "]";  return o; }
vector<int> range(const int x, const int y) { vector<int> v(y - x + 1); iota(v.begin(), v.end(), x); return v; }
template <typename T> istream& operator>>(istream& i, vector<T>& o) { rep(j, o.size()) i >> o[j]; return i;}
template <typename T, typename S, typename U> ostream &operator<<(ostream &o, const priority_queue<T, S, U> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.top(); tmp.pop(); o << x << " ";} return o; }
template <typename T> ostream &operator<<(ostream &o, const queue<T> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.front(); tmp.pop(); o << x << " ";} return o; }
template <typename T> ostream &operator<<(ostream &o, const stack<T> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.top(); tmp.pop(); o << x << " ";} return o; }
template <typename T> unordered_map<T, ll> counter(vector<T> vec){unordered_map<T, ll> ret; for (auto&& x : vec) ret[x]++; return ret;};
string substr(string s, P x) {return s.substr(x.fi, x.se - x.fi); }
void vizGraph(vvll& g, int mode = 0, string filename = "out.png") { ofstream ofs("./out.dot"); ofs << "digraph graph_name {" << endl; set<P> memo; rep(i, g.size())  rep(j, g[i].size()) { if (mode && (memo.count(P(i, g[i][j])) || memo.count(P(g[i][j], i)))) continue; memo.insert(P(i, g[i][j])); ofs << "    " << i << " -> " << g[i][j] << (mode ? " [arrowhead = none]" : "")<< endl;  } ofs << "}" << endl; ofs.close(); system(((string)"dot -T png out.dot >" + filename).c_str()); }
size_t random_seed; namespace std { using argument_type = P; template<> struct hash<argument_type> { size_t operator()(argument_type const& x) const { size_t seed = random_seed; seed ^= hash<ll>{}(x.fi); seed ^= (hash<ll>{}(x.se) << 1); return seed; } }; }; // hash for various class
#define ldout fixed << setprecision(40) 

#define EPS (double)1e-14
#define INF (ll)1e18
#define mo  (ll)(1e9+7)
// end of hamko utils

int main() {
  int n, s, k;
  cin >> n >> s >> k;
  const int mod = 1e9+7;
// i人目が1円多く払うことにするとそれ以降の人全員が最低1円多く払わないといけなくなる
// ∴ クラス全体で見るとn-i円増える
  int dp[101][20020];
	dp[0][0] = 1;
	for(int i = 0; i <n; ++i){
    for(int j = 0; j < s+1; ++j){
      if (i && j-k*(n-i)>=0) (dp[i][j] += dp[i-1][j-k*(n-i)]) %= mod;
  		if (j-(n-i)>=0) (dp[i][j] += dp[i][j-(n-i)]) %= mod;
    }  
  } 
	cout << dp[n-1][s] << endl;
}
0