首先微扰法证明贪心策略，目前假如到了第$i$个人这里，后面的人编号是$i+1$，设左手是$l_i$,右手是$r_i$,然后我们前面部分的结果都是$l_1*l_2*……*l_{i-1}$,我们目前计算第$i+1$这个人的得分，假如不交换顺序，其实就是前面的乘积在乘以$l_i$然后除以$r_{i+1}$,设前部分乘积是$x$,也就是

$\frac {x*l_i}{r_{i+1}}$,交换了就是$\frac {x*l_{i+1}}{r_i}$

如果交换前的结果小就是

$\frac {x*l_i}{r_{i+1}}$ < $\frac {x*l_{i+1}}{r_i}$

通分以后约掉$x$你发现就是

$l_i*r_i<l_{i+1}*r_{i+1}$

所以按照$l_i*r_i$小的优先，排序即可。

然后就是高精度乘法和除法的结合，这里预处理了前缀积的结果到字符串$sum$

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e3 + 5;
int n;
struct node
{
	int l, r;
	bool operator < (const node &x) const
	{
		return l * r < x.l * x.r;
	}
} a[N];
vector<int> s[N];
vector<int> mul(vector<int> &a, int b)
{
	vector<int> c;
	int t = 0;
	for (int i = 0; i < a.size(); i++) 
	{
		t += a[i] * b;
		c.push_back(t % 10);
		t /= 10;
	}
	while (t)
	{
		c.push_back(t % 10);
		t /= 10;
	}
	return c;
}
vector<int> div(vector<int> &a, int b)
{
	vector<int> c;
	int x = 0;
	for (int i = a.size() - 1; i >= 0; i--)
	{
		x = x * 10 + a[i];
		c.push_back(x / b);
		x %= b;
	}
	reverse(c.begin(), c.end());
	while (c.size() > 1 && !c.back()) c.pop_back();
//	reverse(c.begin(), c.end());
	return c;
}
void print(vector<int> &c)
{
	for (int i = c.size() - 1; i >= 0; i--) cout << c[i];
	cout << "\n";
}
vector<int> cmp(vector<int> &a, vector<int> &b)
{
	if (a.size() > b.size()) return a;
	if (b.size() > a.size()) return b;
	for (int i = a.size() - 1; i >= 0; i--)
	{
		if (a[i] > b[i]) return a;
		if (a[i] < b[i]) return b;
	}
	return a;
}
int main()
{   
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n;
	for (int i = 0; i <= n; i++) cin >> a[i].l >> a[i].r;
	sort(a + 1, a + n + 1);
	for (int i = 0; i < n; i++)
	{
		if (i == 0) s[i].push_back(a[0].l);
		else
		{
			s[i] = mul(s[i - 1], a[i].l);
		}
	}
	vector<int> ans(1, 0);
	for (int i = 1; i <= n; i++)
	{
		vector<int> b = div(s[i - 1], a[i].r);
		ans = cmp(ans, b);
	}
	print(ans);
    return 0; 
}

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