Problem:
Score : 100 points

Problem Statement

You are given two integers A and B.
Find the largest value among A+B, A-B and A \times B.

Constraints

-1000 \leq A,B \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the largest value among A+B, A-B and A \times B.

Sample Input 1

3 1

Sample Output 1

4

3+1=4, 3-1=2 and 3 \times 1=3. The largest among them is 4.

Sample Input 2

4 -2

Sample Output 2

6

The largest is 4 - (-2) = 6.

Sample Input 3

0 0

Sample Output 3

0

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <numeric>




typedef long long ll;
#define PI 3.14159265359
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define mp make_pair
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define decp(n) cout<<fixed<<setprecision((int)n)
#define _GLIBCXX_DEBAC
#include <functional>
/*map<Keyの型, Valueの型> 変数名;
queue<型> 変数名;
priority_queue<型> 変数名;
priority_queue<型, vector<型>, greater<型>> 変数名;*/
//copy(v1.begin(), v1.end(), back_inserter(v2));
using namespace std;

ll gcd(ll a, ll b) {
	ll c;
	if (a < b) {
		c = a; a = b; b = c;
	}
	while (b != 0) {
		c = a % b; a = b; b = c;
	}
	return a;
}

ll lcm(ll a, ll b) {
	ll c;
	c = a * b / gcd(a, b);
	return c;
}

int sort_greater(int m, int n) {
	vector<int> a(m);
	for (int i = 0; i < m; i++) {
		cin >> a[i];
	}
	sort(a.begin(), a.end(), greater<int>());
	for (int i = 0; i < n; i++) {
		cout << a[i];
	}
	return 0;
}

int bubblesort(int a[], int n) {
	int sw = 0;
	bool flag = 1;
	for (int i = 0; flag; i++) {
		flag = 0;
		for (int j = n - 1; j >= i + 1; j--) {
			if (a[j] < a[j - 1]) {
				swap(a[j], a[j - 1]);
				flag = 1;
				sw += 1;
			}
		}
	}
	return sw;
}

int a_z()
{
	for (int i = 0; i <= ('Z' - 'A'); i++) {
		cout << (char)('A' + i);
	}
	return 0;
}

int selecttionsort(int a[], int n) {
	int t, sw = 0, minj;
	for (int i = 0; i < n - 1; i++) {
		minj = i;
		for (int j = 0; j < n - 1; j++) {
			if (a[j] < a[minj]) {
				minj = j;
			}
		}
		t = a[i]; a[i] = a[minj]; a[minj] = t;
		if (i != minj) sw += 1;

	}
	return sw;
}

bool is_prime(int x) {
	if (x <= 1) return false;
	for (int i = 2; i * i <= x; i++) {
		if (x % i == 0) return false;
	}
	return true;
}

const ll mod = 1000000007;

map<string, int> memo;

using P=pair<ll,ll>;

/*void dfs(int v, int p = -1) {
	for (int u : to[v]) {
		if (u == p) continue;
		ans[u] += ans[v];
		dfs(u, v);
	}
}*/
ll findIndex(int value, vector<ll> a)
{

	for (ll i = 0; i < (ll)a.size(); i++) {
		if (a[i] == value) {
			return i;
		}
	}
	return -1;
}

struct mint {
	ll x;
	mint(ll x=0):x(x%mod){}
	mint pow(ll t) const{
		if (!t)return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1)a *= *this;
		return a;
	}
	mint inv() const {
		return pow(mod - 2);
	}
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod)x -= mod;
		return *this;
	}
	mint& operator *= (const mint a) {
		(x *= a.x) %= mod;
		return *this;
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a)const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) {
		mint res(*this);
		return res *= a;
	}
	mint operator/(const mint a) {
		mint res(*this);
		return res /= a;
	}
};

struct combination {
	vector<mint> fact, ifact;
	combination(ll n) :fact(n + 1), ifact(n + 1) {
		
		fact[0] = 1;
		for (ll i = 1; i <= n; i++) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (ll i = n; i >= 1; i--) {
			ifact[i - 1] = ifact[i] * i;
		}
	}
	mint operator()(ll n, ll k) {
		if (k<0 || k>n)return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
};



const int INF = 1001001001;
const int M = 1000001;
string f(ll n) {
	if (n == 0)return "";
	n--;
	return f(n / 26) + string(1, 'a' + n % 26);
}

string x(ll n) {
	string s = "";
	if (n == 0)return "";
	n--;
	s += 'a' + n % 26;
	return x(n / 26) + s ;
}

int main() {
	int a, b;
	cin >> a >> b;
	cout << max({ a + b,a - b,a * b });
	return 0;
}