[AcWing 97] 约数之和

分治 递归

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#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;
const LL mod = 9901;

LL a, b;

LL qmi(LL a, LL k)
{
    LL res = 1;
    while (k) {
        if (k & 1)
            res = res * a % mod;
        k >>= 1;
        a = a * a % mod;
    }
    return res;
}

LL sum(LL p, LL k)
{
    if (k == 1)
        return 1;
    if (k % 2)
        return (sum(p, k -1) + qmi(p, k - 1)) % mod;
    else
        return (1 + qmi(p, k / 2)) * sum(p, k / 2) % mod;
}

void solve()
{
    cin >> a >> b;
    LL res = 1;
    for (LL i = 2; i <= a / i; i ++) {
        if (a % i == 0) {
            LL cnt = 0;
            while (a % i == 0) {
                a /= i;
                cnt ++;
            }
            res = res * sum(i, cnt * b + 1) % mod;
        }
    }
    if (a > 1)
        res = res * sum(a, b + 1) % mod;
    if (a == 0)
        res = 0;
    cout << res << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    solve();

    return 0;
}

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1. 约数之和
   由算数基本定理可知，\(A = P_1^{\alpha_1} \cdot P_2^{\alpha_2} \cdot \cdots \cdot P_k^{\alpha_k}\)，那么 \(A^{B} = P_1^{\alpha_1 \cdot B} \cdot P_2^{\alpha_2 \cdot B} \cdot \cdots \cdot P_k^{\alpha_k \cdot B}\)
   对于 \(A\)，约数之和为 \((P_1^{0} + P_1^{1} + \cdots P_1^{\alpha_1}) \cdot (P_2^{0} + P_2^{1} + \cdots P_2^{\alpha_2}) \cdots (P_k^{0} + P_k^{1} + \cdots P_k^{\alpha_k})\)
   对于 \(A^{B}\)，约数之和为 \((P_1^{0} + P_1^{1} + \cdots P_1^{\alpha_1 \cdot B}) \cdot (P_2^{0} + P_2^{1} + \cdots P_2^{\alpha_2 \cdot B}) \cdots (P_k^{0} + P_k^{1} + \cdots P_k^{\alpha_k \cdot B})\)
2. 公式推导
   定义 \(sum(P, k) = P^{0} + P^{1} + \cdots + P^{k - 1}\)
   ① 当 \(k\) 是偶数时
   \(sum(P, k) = P^{0} + P^{1} + \cdots + P^{\frac{k}{2} - 1} + P^{\frac{k}{2}} + P^{\frac{k}{2} + 1} + \cdots + P^{k - 1} = P^{0} + P^{1} + \cdots + P^{\frac{k}{2} - 1} + P^{\frac{k}{2}} \cdot (P^{0} + P^{1} + \cdots + P^{\frac{k}{2} - 1}) = sum(P, \frac{k}{2}) + P^{\frac{k}{2}} \cdot sum(P, \frac{k}{2}) = (1 + P^{\frac{k}{2}}) \cdot sum(P, \frac{k}{2})\)
   ② 当 \(k\) 是奇数时
   \(sum(P, k) = P^{0} + P^{1} + \cdots + P^{k - 2} + P^{k - 1} = sum(P, k - 1) + P^{k - 1}\)，\(k - 1\) 是偶数，可以用偶数的递推公式
3. 特殊情况
   当 \(a = 0\) 时，约数之和为 \(0\)