CF1097D Makoto and a Blackboard

思路：

概率dp。首先对n进行因子分解得到，然后每个素因子p_i，计算经过k次操作之后的期望值E_i，再利用期望的性质把所有E_i乘起来得到最终结果。E_i可以通过概率dp计算，dp[i][j]表示经过i次操作之后出现的概率。

实现：

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const ll MOD = 1e9 + 7;

ll dp[10005][60];

ll qpow(ll x, ll n)
{
    ll res = 1;
    while (n)
    {
        if (n & 1) res = res * x % MOD;
        x = x * x % MOD;
        n >>= 1;
    }
    return res;
}

ll inv(ll x)
{
    return qpow(x, MOD - 2);
}

map<ll, int> fac(ll x)
{
    map<ll, int> ans;
    for (ll i = 2; i * i <= x; i++)
    {
        while (x % i == 0)
        {
            x /= i;
            ans[i]++;
        }
    }
    if (x != 1) ans[x] = 1;
    return ans;
}

int main()
{
    ll n; int k;
    while (cin >> n >> k)
    {
        map<ll, int> ans = fac(n);
        ll res = 1;
        for (auto it: ans)
        {
            memset(dp, 0, sizeof dp);
            ll tmp = it.first; int cnt = it.second;
            dp[0][cnt] = 1;
            for (int i = 1; i <= k; i++)
            {
                dp[i][cnt] = dp[i - 1][cnt] * inv(cnt + 1) % MOD;
                for (int j = cnt - 1; j >= 0; j--)
                {
                    dp[i][j] = dp[i][j + 1];
                    dp[i][j] = (dp[i][j] + dp[i - 1][j] * inv(j + 1)) % MOD;
                }
            }
            ll sum = 0;
            for (int i = 0; i <= cnt; i++)
            {
                sum = (sum + dp[k][i] * qpow(tmp, i) % MOD) % MOD;
            }
            res = res * sum % MOD;
        }
        cout << res << endl;
    }
    return 0;
}