Given an integer n, the task is to find the smallest positive integer N such that the number of divisors of N equals n. This problem leverages the divisor count formula: if N = ∏ p_i^{k_i}, then the number of divisors is ∏ (k_i + 1). Thus, n = ∏ (k_i + 1), where k_i are non-negative integers.

A naive approach involves factorizing n into its divisors (k_i + 1) and exploring all combinations to compute N = ∏ p_i^{k_i}, where p_i are consecutive primes. This can be done via depth-first search (DFS) over the factors of n, but direct computation of large integers (using big integer arithmetic) for each candidate N leads to high time complexity, approximately O(n√n).

To optimize, logarithmic properties are uitlized: log(xy) = log(x) + log(y) and log(x^y) = y log(x). Instead of comparing large integers driectly during DFS, compare their logarithms. This reduces computational overhead since logarithmic values are precomputed and stored.

Implementation steps:

1. Precompute primes up to a sufficient limit (e.g., 600) and their natural logarithms.
2. For each integer up to n, precompute its divisors to speed up DFS.
3. Define a structure to store exponents k_i for primes, with a comparison operator based on logarithmic sums.
4. Use DFS to iterate over factorizations of n, updating the minimal logarithmic sum.
5. After DFS, compute the final integer N using fast exponentiation with the optimal exponents.

This optimization reduces runtime significant, e.g., from 700 ms to 20 ms for n = 50000.

Code example with restructured logic and renamed variables:

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

struct BigInt {
    vector<int> digits;
    bool isNegative;
    BigInt() : digits(1, 0), isNegative(false) {}
    BigInt(long long x) {
        if (x < 0) { isNegative = true; x = -x; }
        else isNegative = false;
        digits.clear();
        do {
            digits.push_back(x % 10);
            x /= 10;
        } while (x > 0);
    }
    bool operator<(const BigInt& other) const {
        if (digits.size() != other.digits.size()) return digits.size() < other.digits.size();
        for (int i = digits.size() - 1; i >= 0; --i)
            if (digits[i] != other.digits[i]) return digits[i] < other.digits[i];
        return false;
    }
    BigInt operator*(const BigInt& other) const {
        BigInt result;
        result.digits.assign(digits.size() + other.digits.size(), 0);
        for (size_t i = 0; i < digits.size(); ++i)
            for (size_t j = 0; j < other.digits.size(); ++j) {
                result.digits[i + j] += digits[i] * other.digits[j];
                result.digits[i + j + 1] += result.digits[i + j] / 10;
                result.digits[i + j] %= 10;
            }
        while (result.digits.size() > 1 && result.digits.back() == 0) result.digits.pop_back();
        result.isNegative = isNegative ^ other.isNegative;
        return result;
    }
    BigInt pow(int exp) const {
        if (exp == 0) return BigInt(1);
        BigInt half = pow(exp / 2);
        half = half * half;
        if (exp % 2) half = half * (*this);
        return half;
    }
};

const int MAX_PRIME = 600;
vector<int> primes;
vector<double> logPrimes;
vector<vector<int>> divisors;

void sieve() {
    vector<bool> isPrime(MAX_PRIME + 1, true);
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i * i <= MAX_PRIME; ++i)
        if (isPrime[i])
            for (int j = i * i; j <= MAX_PRIME; j += i)
                isPrime[j] = false;
    for (int i = 2; i <= MAX_PRIME; ++i)
        if (isPrime[i]) {
            primes.push_back(i);
            logPrimes.push_back(log(i));
        }
}

void precomputeDivisors(int n) {
    divisors.resize(n + 1);
    for (int i = 1; i <= n; ++i) {
        for (int d = 2; d * d <= i; ++d) {
            if (i % d == 0) {
                divisors[i].push_back(d);
                if (d != i / d) divisors[i].push_back(i / d);
            }
        }
        divisors[i].push_back(i);
    }
}

struct ExponentState {
    vector<int> exponents;
    bool operator<(const ExponentState& other) const {
        double sum1 = 0, sum2 = 0;
        for (size_t i = 0; i < exponents.size(); ++i) {
            sum1 += exponents[i] * logPrimes[i];
            sum2 += other.exponents[i] * logPrimes[i];
        }
        return sum1 < sum2;
    }
    void adjust(int idx, int delta) { exponents[idx] += delta; }
};

ExponentState bestState;

void dfs(int remaining, ExponentState current, int primeIdx) {
    if (bestState < current) return;
    if (remaining == 1) {
        if (current < bestState) bestState = current;
        return;
    }
    for (int factor : divisors[remaining]) {
        current.adjust(primeIdx, factor - 1);
        dfs(remaining / factor, current, primeIdx + 1);
        current.adjust(primeIdx, -(factor - 1));
    }
}

int main() {
    int n;
    cin >> n;
    sieve();
    precomputeDivisors(n);
    bestState.exponents.assign(primes.size(), 10000);
    ExponentState start;
    start.exponents.assign(primes.size(), 0);
    dfs(n, start, 0);
    BigInt result(1);
    for (size_t i = 0; i < primes.size(); ++i) {
        if (bestState.exponents[i] == 0) break;
        result = result * BigInt(primes[i]).pow(bestState.exponents[i]);
    }
    for (int i = result.digits.size() - 1; i >= 0; --i) cout << result.digits[i];
    cout << endl;
    return 0;
}