結果

問題 No.2896 Monotonic Prime Factors
ユーザー 👑 binapbinap
提出日時 2024-09-20 23:01:48
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 377 ms / 2,000 ms
コード長 4,595 bytes
コンパイル時間 4,795 ms
コンパイル使用メモリ 273,876 KB
実行使用メモリ 22,320 KB
最終ジャッジ日時 2024-09-20 23:01:58
合計ジャッジ時間 8,738 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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 1 ms
5,248 KB
testcase_01 AC 2 ms
5,248 KB
testcase_02 AC 2 ms
5,248 KB
testcase_03 AC 2 ms
5,248 KB
testcase_04 AC 261 ms
5,376 KB
testcase_05 AC 377 ms
22,320 KB
testcase_06 AC 333 ms
14,640 KB
testcase_07 AC 282 ms
5,376 KB
testcase_08 AC 274 ms
5,376 KB
testcase_09 AC 377 ms
22,232 KB
testcase_10 AC 149 ms
5,732 KB
testcase_11 AC 36 ms
5,376 KB
testcase_12 AC 20 ms
5,376 KB
testcase_13 AC 180 ms
5,848 KB
testcase_14 AC 209 ms
6,492 KB
testcase_15 AC 10 ms
5,376 KB
testcase_16 AC 71 ms
5,376 KB
testcase_17 AC 31 ms
5,376 KB
testcase_18 AC 268 ms
7,768 KB
testcase_19 AC 36 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 # 

#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;

ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
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<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);}

// Sieve of Eratosthenes
// https://youtu.be/UTVg7wzMWQc?t=2774
struct Sieve {
  int n;
  vector<int> f, primes;
  Sieve(int n=1):n(n), f(n+1) {
    f[0] = f[1] = -1;
    for (long long i = 2; i <= n; ++i) {
      if (f[i]) continue;
      primes.push_back(i);
      f[i] = i;
      for (long long j = i*i; j <= n; j += i) {
        if (!f[j]) f[j] = i;
      }
    }
  }
  bool isPrime(int x) { return f[x] == x;}
  vector<int> factorList(int x) {
    vector<int> res;
    while (x != 1) {
      res.push_back(f[x]);
      x /= f[x];
    }
    return res;
  }
  vector<pair<ll,int>> factor(ll x) {
    vector<pair<ll,int>> res;
    for (int p : primes) {
      int y = 0;
      while (x%p == 0) x /= p, ++y;
      if (y != 0) res.emplace_back(p,y);
    }
    if (x != 1) res.emplace_back(x,1);
    return res;
  }
	int num_of_divisors(long long x){
		auto f = factor(x);
		long long res = 1;
		for(auto [p, val] : f){
			res *= val + 1;
		}
		return res;
	}
	set<long long> divisors(long long x){
		set<long long> se = {1};
		auto f = factor(x);
		for(auto [key, value] : f){
			set<long long> old = se;
			for(long long y : old){
				rep(i, value + 1){
					se.insert(y);
					y *= key;
				}
			}
		}
		return se;
	}
};


using mint = modint998244353;

// combination mod prime
// https://youtu.be/8uowVvQ_-Mo?t=6002
// https://youtu.be/Tgd_zLfRZOQ?t=9928
struct modinv {
  int n; vector<mint> d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vector<mint> d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*n), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} facts;
struct modfactinv {
  int n; vector<mint> d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*invs(n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} ifacts;
mint comb(int n, int k) {
  if (n < k || k < 0) return 0;
  return facts(n)*ifacts(k)*ifacts(n-k);
}

Sieve sieve(350);


int main(){
	int q;
	cin >> q;
	int x = 0;
	rep(i, q){
		int a, b;
		cin >> a >> b;
		auto f = sieve.factor(a);
		for(auto [prime, time] : f){
			x += time;
		}
		if(x < b){
			cout << "0\n";
		}else{
			mint ans = comb(x - 1, b- 1);
			cout << ans << "\n";
		}
	}
	
	return 0;
}
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