#include<iostream>
#define x first
#define y second
using namespace std;
const int N = 10010;
typedef pair<int, int> PII;
PII p[N];
int n;
bool check01(int mid)
{
    for (int i = 1; i <= n; ++ i) if ((p[i].x / mid) > p[i].y) return false;
    return true;
}

bool check02(int mid)
{
    for (int i = 1; i <= n; ++ i) if ((p[i].x / mid) < p[i].y) return false;
    return true;
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; ++ i) scanf("%d%d", &p[i].x, &p[i].y);

    int l = 1, r = 1e9;
    while (l < r)
    {
        int mid = (l + r) >> 1;
        if (check01(mid)) r = mid;
        else l = mid + 1;
    }
    cout << r << ' ';
    l = 1, r = 1e9;
    while (l < r)
    {
        int mid = (l + r + 1) >> 1;
        if (check02(mid)) l = mid;
        else r = mid - 1;
    }
    cout << l << endl;
    return 0;
}

设答案为 $v$, 因为能炼出 $b$ 个但炼不出 $b+1$ 个所以有 $b×v≤a$ 且 $a<(b+1)×v$ 即 $\frac{a}{b+1}<v⩽\frac a b$

最小值 $\frac{a}{b+1}+1$，最大值 $\frac a b$

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int main()
{
    int n;
    scanf("%d", &n);
    int minn = 1, maxx = 1e9;
    while (n -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);
        minn = max(minn, a / (b + 1) + 1);
        maxx = min(maxx, a / b);
    }

    printf("%d %d\n", minn, maxx);
    return 0;
}