60分代码，4个TLE，我是枚举的质数次数，请问除了换成枚举质数（第一篇题解）如何才能提高搜索效率呢？

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5 * 1e4 + 1919, N = 5 * 1e4;
typedef long long LL;
const LL INF = 2000000000ll;
int prime[MAXN], v[MAXN], num;
LL S;
LL power[MAXN][39], ans[MAXN];
int max_pow[MAXN], c[MAXN];
void prepare() {
    for (int i = 2; i <= N; i++) {
        if (!v[i]) {
            v[i] = i;
            prime[++num] = i;
        }
        for (int j = 1; j <= num; j++) {
            if (prime[j] > v[i] || prime[j] * i > N)
                break;
            v[prime[j] * i] = prime[j];
        }
    }
    for (int i = 1; i <= num; i++) {
        int p = prime[i];
        power[i][0] = 1;
        for (int j = 1; j <= 31; j++) {
            power[i][j] = power[i][j - 1] * p;
            if (power[i][j - 1] * p > INF) {
                max_pow[i] = j - 1;
                break;
            }
        }
    }
}
inline bool pd(LL x) {
    if (x == 1)
        return false;
    for (LL i = 2; i * i <= x; i++)
        if (x % i == 0)
            return false;
    return true;
}
inline LL calc(int t, int ci) { return (power[t][ci + 1] - 1) / (prime[t] - 1); }
int m;
void dfs(int t, LL now, LL now_ans) {
    if (now == 1ll) {
        ans[++m] = now_ans;
        return;
    }
    if (t == num) {
        if (pd(now - 1) && now > prime[t]) {
            ans[++m] = now_ans * (now - 1);
        }
        return;
    }
    for (int i = 0; i <= max_pow[t]; i++) {
        LL cal = calc(t, i);
        if (cal > now) break;
        if (now % cal == 0) {
            c[t] = i;
            dfs(t + 1, now / cal, now_ans * power[t][c[t]]);
            c[t] = 0;
        }
    }
}
inline void solve() {
    m = 0;
    dfs(1, S, 1);
    sort(ans + 1, ans + m + 1);
    cout << m << endl;
    if (m) {
        for (int i = 1; i <= m; i++) printf("%lld ", ans[i]);
        printf("\n");
    }
}
int main() {
    prepare();
    while (scanf("%lld", &S) == 1) {
        solve();
    }
    return 0;
}