A、Groundhog and 2-Power Representation

Python的签到题，会Python就很简单，直接替换加eval就行了

s = input()
s = s.replace('(', '**(')
print(eval(s))

F、Groundhog Looking Dowdy

一共有n天，我们要选择其中m天。每天必须穿一件衣服，没有上限控制，后面有n行，第一个数代表当天衣服数目，紧接着就是每件衣服的权值。
我们要选出m天中，最大减掉最小权值的最小值是多少？
最小值，考虑二分答案求解。
首先使用二元组，第一元是衣服的权值，第二元是第几天能穿。
按照sort默认的二元组进行升序排序，对着这个二元组进行二分操作。直接二分的就是答案。
再使用双游标，去累计是否存在一个区间满足条件最大值减掉最小值有m天的情况，如果存在return true。
时间复杂度O(n*logn)

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 2e6 + 7;
vector<pai> p;
int vis[N], n, m;

bool check(int x) {
    int l = 0, cnt = 0;
    ms(vis, 0);
    for (auto it : p) {
        int val = it.first, day = it.second;
        if (val <= x + p[l].first) {
            ++vis[day];
            if (vis[day] == 1)    ++cnt;
            if (cnt >= m)    return true;
        }
        else {
            while (val > x + p[l].first) {
                --vis[p[l].second];
                if (vis[p[l].second] == 0)    --cnt;
                ++l;
            }
            ++vis[day];
            if (vis[day] == 1)    ++cnt;
            if (cnt >= m)    return true;
        }
    }
    return false;
}

int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; ++i) {
        int num = read();
        for (int j = 1; j <= num; ++j) {
            int x = read();
            p.emplace_back(x, i);
        }
    }
    sort(all(p));
    p.erase(unique(all(p)), p.end());
    int l = 0, r = 1e9, ans;
    while (l <= r) {
        int mid = l + r >> 1;
        if (check(mid))
            ans = mid, r = mid - 1;
        else
            l = mid + 1;
    }
    print(ans);
    return 0;
}

I、The Crime-solving Plan of Groundhog

观察可得，题目所求最小值，一定是一个最小非零一位数乘以其他数构成得合法得最小多位数得积。
接下来就只要大数模拟就行了，前提你的大数常数一定要小，本地都跑半天的，建议直接上模拟乘法。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

// base and base_digits must be consistent
constexpr int base = 1000000000;

constexpr int base_digits = 9;
struct bigint
{
    // value == 0 is represented by empty z
    vector<int> z; // digits

    // sign == 1 <==> value >= 0
    // sign == -1 <==> value < 0
    int sign;

    bigint() : sign(1) {}

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

    bigint& operator=(long long v)
    {
        sign = v < 0 ? -1 : 1;
        v *= sign;
        z.clear();
        for (; v > 0; v = v / base)
            z.push_back((int)(v % base));
        return *this;
    }

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

    bigint& operator+=(const bigint& other)
    {
        if (sign == other.sign)
        {
            for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
            {
                if (i == z.size())
                    z.push_back(0);
                z[i] += carry + (i < other.z.size() ? other.z[i] : 0);
                carry = z[i] >= base;
                if (carry)
                    z[i] -= base;
            }
        }
        else if (other != 0 /* prevent infinite loop */)
        {
            *this -= -other;
        }
        return *this;
    }

    friend bigint operator+(bigint a, const bigint& b)
    {
        return a += b;
    }

    bigint& operator-=(const bigint& other)
    {
        if (sign == other.sign)
        {
            if (sign == 1 && *this >= other || sign == -1 && *this <= other)
            {
                for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
                {
                    z[i] -= carry + (i < other.z.size() ? other.z[i] : 0);
                    carry = z[i] < 0;
                    if (carry)
                        z[i] += base;
                }
                trim();
            }
            else
            {
                *this = other - *this;
                this->sign = -this->sign;
            }
        }
        else
        {
            *this += -other;
        }
        return *this;
    }

    friend bigint operator-(bigint a, const bigint& b)
    {
        return a -= b;
    }

    bigint& operator*=(int v)
    {
        if (v < 0)
            sign = -sign, v = -v;
        for (int i = 0, carry = 0; i < z.size() || carry; ++i)
        {
            if (i == z.size())
                z.push_back(0);
            long long cur = (long long)z[i] * v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
        }
        trim();
        return *this;
    }

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

    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 = (int)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 = (int)(((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 { 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 = (int)(((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;
    }

    bigint& 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();
        return *this;
    }

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

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

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

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

    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 = (int)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();
    }

    friend bigint operator-(bigint v)
    {
        if (!v.z.empty())
            v.sign = -v.sign;
        return v;
    }

    bigint abs() const
    {
        return sign == 1 ? *this : -*this;
    }

    long long longValue() const
    {
        long long res = 0;
        for (int i = (int)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 < s.size() && (s[pos] == '-' || s[pos] == '+'))
        {
            if (s[pos] == '-')
                sign = -sign;
            ++pos;
        }
        for (int i = (int)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 < p.size(); i++)
            p[i] = p[i - 1] * 10;
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for (int v : a)
        {
            cur += v * 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 < a1b1.size(); i++)
            r[i] -= a1b1[i];
        for (int i = 0; i < a2b2.size(); i++)
            r[i] -= a2b2[i];

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

const int N = 1e5 + 7;
string s;
int a[N];

int main() {
    js;
    int T;    cin >> T;
    while (T--) {
        int n;    cin >> n;
        for (int i = 1; i <= n; ++i)    cin >> a[i];
        sort(a + 1, a + 1 + n);
        bigint c = 0, d = 0;
        int pos = -1;
        for (int i = 1; i <= n; ++i) {
            if (a[i] != 0) {
                c = a[i], pos = i;
                break;
            }
        }
        s = char(a[pos + 1] + '0');
        d = a[pos + 1];
        for (int i = 1; i <= n; ++i) {
            if (i == pos or i == pos + 1)    continue;
            char tmp = a[i] + '0';
            s += tmp;
        }
        d.read(s);

        cout << c * d << endl;
    }
    return 0;
}