p473.ans1
======================
6

=================
p473.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int K;

bigint pow_int(int base, int exp) {
    if(exp == 0) {
        return bigint(1);
    }

    if(exp <= 30) {
        bigint res = 1;
        for(int i = 0; i < exp; i++) {
            res *= base;
        }
        return res;
    }

    bigint res = 1;
    bigint b = base;
    for(; exp; exp >>= 1) {
        if(exp & 1) {
            res *= b;
        }
        if(exp > 1) {
            b *= b;
        }
    }
    return res;
}

bool try_u64(
    const vector<int>& factors, const vector<int>& primes, uint64_t& out
) {
    __int128 n = 1;
    for(int i = 0; i < (int)factors.size(); i++) {
        __int128 p = primes[i];
        __int128 pe = 1;
        for(int e = 0; e < factors[i] - 1; e++) {
            pe *= p;
            if(pe > (__int128)UINT64_MAX) {
                return false;
            }
        }
        n *= pe;
        if(n > (__int128)UINT64_MAX) {
            return false;
        }
    }
    out = (uint64_t)n;
    return true;
}

void read() { cin >> K; }

void solve() {
    // The number of "variants of evolution" of N equals its number of divisors.
    // For N = p1^a1 * ... * pm^am that count is (a1+1) * ... * (am+1), so we
    // need the smallest N with exactly K divisors.
    //
    // Every way to write K = b1 * b2 * ... * bm with each bi >= 2 corresponds
    // to a candidate N = p1^(b1-1) * ... * pm^(bm-1), where pi is the i-th
    // prime. To minimize N the larger exponents belong to the smaller primes,
    // so we generate factorizations with non-increasing factors and assign
    // factor i (its exponent being bi - 1) to the i-th prime. We enumerate all
    // such factorizations of K by recursion and take the minimum product, using
    // bigint because K being prime forces the answer 2^(K-1) which is huge.
    //
    // K = 1 means N has a single divisor, so N = 1.

    if(K == 1) {
        cout << 1 << '\n';
        return;
    }

    int max_primes = 0;
    for(int t = K; t > 1; t /= 2) {
        max_primes++;
    }

    vector<int> primes;
    for(int cand = 2; (int)primes.size() < max_primes; cand++) {
        bool is_prime = true;
        for(int p: primes) {
            if(p * p > cand) {
                break;
            }
            if(cand % p == 0) {
                is_prime = false;
                break;
            }
        }

        if(is_prime) {
            primes.push_back(cand);
        }
    }

    vector<double> log_primes(primes.size());
    for(int i = 0; i < (int)primes.size(); i++) {
        log_primes[i] = log((double)primes[i]);
    }

    bigint best;
    bool have_best = false;
    double log_best = numeric_limits<double>::infinity();
    vector<int> factors;

    function<void(int, int)> gen = [&](int remaining, int max_factor) {
        if(remaining == 1) {
            double log_n = 0;
            for(int i = 0; i < (int)factors.size(); i++) {
                log_n += (factors[i] - 1) * log_primes[i];
            }

            if(have_best && log_n > log_best) {
                return;
            }

            uint64_t u;
            if(try_u64(factors, primes, u)) {
                bigint n(to_string(u));
                if(!have_best || n < best) {
                    best = n;
                    have_best = true;
                    log_best = log_n;
                }
                return;
            }

            bigint n = 1;
            for(int i = 0; i < (int)factors.size(); i++) {
                n *= pow_int(primes[i], factors[i] - 1);
                if(have_best && n >= best) {
                    break;
                }
            }

            if(!have_best || n < best) {
                best = n;
                have_best = true;
                log_best = log_n;
            }

            return;
        }

        for(int f = min(remaining, max_factor); f >= 2; f--) {
            if(remaining % f == 0) {
                factors.push_back(f);
                gen(remaining / f, f);
                factors.pop_back();
            }
        }
    };

    gen(K, K);

    cout << best << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}

=================
p473.in1
======================
4

=================
p473.py
======================
import sys

sys.set_int_max_str_digits(100000)


def solve(K):
    # Find the smallest positive integer N such that N has exactly K divisors.
    #
    # The key insight is that if N = p1^a1 * p2^a2 * ... * pm^am, then
    # the number of divisors is (a1+1) * (a2+1) * ... * (am+1).
    #
    # So we need to find all factorizations of K = b1 * b2 * ... * bm where
    # each bi >= 2, and then N = p1^(b1-1) * p2^(b2-1) * ... * pm^(bm-1)
    # where p1, p2, ..., pm are the first m primes.
    #
    # We use Python for potentially large integers when K is prime, as the
    # answer would be 2^(K-1) which can be astronomically large.
    #
    # This approach is fast because there aren't that many factorizations
    # of a number <= 10^5. In particular, an easy upper bound is the log(K)-th Bell 
    # number equal to ~1e9 for 16. However, this is only possible for K = 2^c, in 
    # which case the factors are the same and if we don't generate them multiple times
    # we significantly reduce the above figure. For example if K = 2^c a tighter upper
    # bound for the number of unique factorizations is only 2^(c-1) - think of this as 
    # setting the separators between the groups.

    if K == 1:
        return 1

    # Generate enough primes (we won't need more than log2(K) primes)
    # since the smallest factorization would be 2 * 2 * ... * 2
    max_primes_needed = K.bit_length()
    primes = []
    is_prime = [True] * (max_primes_needed * 10)  # Generous upper bound

    for i in range(2, len(is_prime)):
        if is_prime[i]:
            primes.append(i)
            if len(primes) >= max_primes_needed:
                break
            for j in range(i * i, len(is_prime), i):
                is_prime[j] = False

    # Find all ordered factorizations of K using iterative approach
    # Each factorization represents exponents+1 for our prime factorization
    factorizations = []

    # Stack: (remaining_value, current_factorization, min_factor)
    stack = [(K, [], 2)]

    while stack:
        remaining, current, min_factor = stack.pop()

        if remaining == 1:
            factorizations.append(current[::-1])
            continue

        # Try all factors from min_factor to remaining
        # This ensures we generate factorizations in non-increasing order
        for factor in range(min_factor, remaining + 1):
            q, r = divmod(remaining, factor)
            if r == 0:
                new_current = current + [factor]
                stack.append((q, new_current, factor))

    # For each factorization, compute the corresponding N
    min_n = None

    for factors in factorizations:
        # factors = [b1, b2, ..., bm] where K = b1 * b2 * ... * bm
        # N = p1^(b1-1) * p2^(b2-1) * ... * pm^(bm-1)

        assert len(factors) <= len(primes)

        n = 1
        for i, factor in enumerate(factors):
            n *= primes[i] ** (factor - 1)

            # Early termination if n is already too large
            if min_n is not None and n >= min_n:
                break

        if min_n is None or n < min_n:
            min_n = n

    return min_n


def main():
    K = int(input().strip())
    print(solve(K))


if __name__ == "__main__":
    main()

=================
statement.txt
======================
473. Droid formation
Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



A long time ago (but somehow in the future), in a Galaxy Far Far Away...

— Your majesty, Jedi detachment almost finished to mine our new Death Cube! New battle droids are unable to restrain them! What to do!?

— Rest assured! What is the strength of every troop of droids?

— 12 droids, your majesty!

— Fools! I told you that a troop should have 4 variants of evolution but troop of 12 droids has 6! This perverts threads of the Power — and infeeds Jedis! Regroup the army — and Jedis will lose!

— Yes sir!

Number of variants of evolution of a troop of droids is the number of ways to draw it up in rows so that every row has the same number of droids. For example, a troop of 12 droids can be arranged in 1 row of 12 droids, 2 rows of 6 droids, 3 rows of 4 droids, 4 rows of 3 droids, 6 rows of 2 droids and 12 rows consisting of 1 droid each. So, as the Emperor noticed, there are 6 variants of evolution for this troop of droids.

You problem is more general — given the number K of favorable variants of evolution, find the smallest positive size of a troop of droids N which has this very number of variants of evolution.

Input
Input file contains only number K from the problem statement (1 ≤ K ≤ 105).

Output
Write to the output file the required number N. If there is no such number, write to the output file number 0 instead.

Example(s)
sample input
sample output
4
6

=================
