結果

問題 No.2896 Monotonic Prime Factors
コンテスト
ユーザー ルク
提出日時 2024-09-20 22:08:19
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
RE  
実行時間 -
コード長 3,618 bytes
コンパイル時間 22,602 ms
コンパイル使用メモリ 357,996 KB
最終ジャッジ日時 2025-02-24 10:14:07
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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 * 15 RE * 3
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 # 

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
#include <chrono>
#include <unistd.h>
using namespace std;
using namespace chrono;
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rep1(i, n) for (ll i = 1; i <= (n); ++i)
#define rrep(i, n) for (ll i = n; i > 0; --i)
#define bitrep(i, n) for (ll i = 0; i < (1 << n); ++i)
#define all(a) (a).begin(), (a).end()
#define yesNo(b) ((b) ? "Yes" : "No")
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using mint = modint998244353;
using MINT = modint1000000007;
string alphabet = "abcdefghijklmnopqrstuvwxyz";
string ALPHABET = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
constexpr double pi = 3.141592653589793;
constexpr ll smallMOD = 998244353;
constexpr ll bigMOD = 1000000007;
constexpr ll dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
constexpr ll dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
struct Init
{
    Init()
    {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout << fixed << setprecision(15);
    }
} init;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec)
{
    os << "[";
    rep(i, vec.size())
    {
        os << vec[i];
        if (i != vec.size() - 1)
            os << ", ";
    }
    os << "]";
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<vector<T>> &vec)
{
    os << "[";
    rep(i, vec.size())
    {
        os << vec[i];
        if (i != vec.size() - 1)
            os << ", ";
    }
    os << "]";
    return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &pair_var)
{
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &st)
{
    os << "{";
    for (auto itr = st.begin(); itr != st.end(); ++itr)
    {
        os << *itr;
        if (next(itr) != st.end())
            os << ", ";
    }
    os << "}";
    return os;
}

void prime_factorize(ll N, vector<pair<ll, ll>> &res)
{
    for (ll p = 2; p * p <= N; ++p)
    {
        if (N % p != 0)
            continue;
        ll e = 0;
        while (N % p == 0)
        {
            ++e;
            N /= p;
        }
        res.emplace_back(p, e);
    }
    if (N != 1)
    {
        res.emplace_back(N, 1);
    }
}

struct NCR
{
    ll max = 500020, mod = 0;
    vector<ll> fact, inv, inv_fact;
    NCR(ll mod = 998244353)
    {
        this->mod = mod;
        fact.resize(max);
        inv.resize(max);
        inv_fact.resize(max);
        fact[0] = 1;
        fact[1] = 1;
        inv[0] = 1;
        inv[1] = 1;
        inv_fact[0] = 1;
        inv_fact[1] = 1;
        for (ll i = 2; i < max; i++)
        {
            fact[i] = fact[i - 1] * i % mod;
            inv[i] = mod - inv[mod % i] * (mod / i) % mod;
            inv_fact[i] = inv_fact[i - 1] * inv[i] % mod;
        }
    }
    ll nCr(ll n, ll r)
    {
        if (n < r)
            return 0;
        if (n < 0 || r < 0)
            return 0;
        ll x = fact[n];
        ll y = inv_fact[n - r];
        ll z = inv_fact[r];
        return x * ((y * z) % mod) % mod;
    }
};

ll sm = 0;
NCR ncr;
void f(ll b)
{
    cout << ncr.nCr(sm - 1, b - 1) << endl;
}

int main()
{
    ll n;
    cin >> n;
    rep(i, n)
    {
        ll a, b;
        cin >> a >> b;
        vector<pair<ll, ll>> res;
        prime_factorize(a, res);
        for (auto p : res)
        {
            sm += p.second;
        }
        f(b);
    }
    return 0;
}
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