#357251 (C++17) No.847 Divisors of Power

• 問題: No.847 Divisors of Power
• ユーザー: kimiyuki
• 提出日時: 2019-07-05 22:59:06
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: TLE
• 実行時間: -
• コード長: 2,634 bytes
• コンパイル時間: 2,275 ms
• コンパイル使用メモリ: 205,892 KB
• 最終ジャッジ日時: 2025-01-07 06:06:18
• ジャッジサーバーID （参考情報）: judge3 / judge1

ファイルパターン	結果
sample	AC * 4
other	AC * 24 TLE * 2

ソースコード

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#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = (int)(n) - 1; (i) >= 0; -- (i))
#define REP3R(i, m, n) for (int i = (int)(n) - 1; (i) >= (int)(m); -- (i))
#define ALL(x) begin(x), end(x)
using ll = long long;
using namespace std;
/**
 * @note O(\sqrt{n})
 * @note about 1.0 sec for 10^5 queries with n < 10^10
 */
struct prepared_primes {
    int size;
    vector<int> sieve;
    vector<int> primes;
    prepared_primes(int size_)
        : size(size_) {
        sieve.resize(size);
        REP3 (p, 2, size) if (sieve[p] == 0) {
            primes.push_back(p);
            for (int k = p; k < size; k += p) {
                if (sieve[k] == 0) {
                    sieve[k] = p;
                }
            }
        }
    }
    vector<ll> prime_factorize(ll n) {
        assert (1 <= n and n < (ll)size * size);
        vector<ll> result;
        // trial division for large part
        for (int p : primes) {
            if (n < size or n < (ll)p * p) {
                break;
            }
            while (n % p == 0) {
                n /= p;
                result.push_back(p);
            }
        }
        // small part
        if (n == 1) {
            // nop
        } else if (n < size) {
            while (n != 1) {
                result.push_back(sieve[n]);
                n /= sieve[n];
            }
        } else {
            result.push_back(n);
        }
        assert (is_sorted(ALL(result)));
        return result;
    }
    vector<ll> list_all_factors(ll n) {
        auto p = prime_factorize(n);
        vector<ll> d;
        d.push_back(1);
        for (int l = 0; l < p.size(); ) {
            int r = l + 1;
            while (r < p.size() and p[r] == p[l]) ++ r;
            int n = d.size();
            REP (k1, r - l) {
                REP (k2, n) {
                    d.push_back(d[d.size() - n] * p[l]);
                }
            }
            l = r;
        }
        return d;
    }
};
int solve(int n, int k, int m) {
    prepared_primes primes(sqrt(n) + 100);
    auto factors = primes.list_all_factors(n);
    unordered_set<int> cur, prv;
    cur.insert(1);
    while (k -- and cur.size() != prv.size()) {
        cur.swap(prv);
        cur.clear();
        for (int d1 : prv) for (int d2 : factors) {
            if ((ll)d1 * d2 <= m) {
                cur.insert(d1 * d2);
            }
        }
    }
    return cur.size();
}
int main() {
    int n, k, m; cin >> n >> k >> m;
    cout << solve(n, k, m) << endl;
    return 0;
}
 
 
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