#338102 (C++14) No.811 約数の個数の最大化

• 問題: No.811 約数の個数の最大化
• ユーザー: fine
• 提出日時: 2019-04-13 00:14:26
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 6 ms / 2,000 ms
• コード長: 2,049 bytes
• コンパイル時間: 1,873 ms
• コンパイル使用メモリ: 178,324 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-09-15 06:29:02
• 合計ジャッジ時間: 2,325 ms
• ジャッジサーバーID （参考情報）: judge2 / judge6

ファイルパターン	結果
sample	AC * 3
other	AC * 12

ソースコード

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
bool is_prime(int n) {
    if (n % 2 == 0) return false;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return n != 1;
}
vector<int> divisor(int n) {
    vector<int> res;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            res.emplace_back(i);
            if (i != n / i) res.emplace_back(n / i);
        }
    }
    return res;
}
map<int, int> prime_factor(int n) {
    map<int, int> res;
    for (int i = 2; i * i <= n; i++) {
        while (n % i == 0) {
            ++res[i];
            n /= i;
        }
    }
    if (n != 1) res[n] = 1;
    return res;
}
vector<int> sieve(int n) {
    vector<int> res;
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = false;
    is_prime[1] = false;
    for (int i = 2; i <= n; i++) {
        if (!is_prime[i]) continue;
        res.emplace_back(i);
        for (int j = 2 * i; j <= n; j += i) is_prime[j] = false;
    }
    return res;
}
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, K;
    cin >> n >> K;
    vector<int> is_prime(n, true);
    is_prime[0] = is_prime[1] = false;
    vector<int> cnt1(n, 1);
    vector<int> cnt2(n, 0);
    int m = n;
    for (int i = 2; i < n; i++) {
        if (!is_prime[i]) continue;
        int tmp = 0;
        while (m % i == 0) {
            tmp++;
            m /= i;
        }
        cnt1[i] = 2;
        cnt2[i] = min(1, tmp);
        for (int j = 2 * i; j < n; j += i) {
            is_prime[j] = false;
            int x = j;
            int tmp2 = 0;
            while (x % i == 0) {
                x /= i;
                tmp2++;
            }
            cnt1[j] *= tmp2 + 1;
            cnt2[j] += min(tmp, tmp2);
        }
    }
    int max_num = 0;
    int ans = 0;
    for (int i = 1; i < n; i++) {
        if (cnt2[i] < K) continue;
        if (cnt1[i] > max_num) {
            max_num = cnt1[i];
            ans = i;
        }
    }
    cout << ans << endl;
    return 0;
}
 
 
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