結果

問題 No.2944 Sigma Partition Problem
ユーザー 👑 binap
提出日時 2024-10-18 22:51:23
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 782 ms / 4,000 ms
コード長 2,678 bytes
コンパイル時間 4,275 ms
コンパイル使用メモリ 254,388 KB
最終ジャッジ日時 2025-02-24 21:05:46
ジャッジサーバーID
（参考情報） 		judge4 / judge
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 * 24
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<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;

template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}

using mint = modint998244353;

const int M = 4000;

int main(){
	int q;
	cin >> q;
	vector<vector<mint>> f(M + 1, vector<mint>(M + 1));
	for(int k = 0; k <= M; k++) f[0][k] = 1;
	for(int n = 1; n <= M; n++){
		for(int k = 1; k <= M; k++){
			for(int m = n; m >= 0; m -= k){
				f[n][k] += f[m][k - 1];
			}
		}
	}
	rep(_, q){
		int t, n, k;
		cin >> t >> n >> k;
		cout << f[n][k] << "\n";
	}
	return 0;
}
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