B - Two Cakes CodeForces - 911B

It’s New Year’s Eve soon, so Ivan decided it’s high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into a pieces, and the second one — into b pieces.

Ivan knows that there will be n people at the celebration (including himself), so Ivan has set n plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:

Each piece of each cake is put on some plate;
Each plate contains at least one piece of cake;
No plate contains pieces of both cakes.
To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number x such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least x pieces of cake.

Help Ivan to calculate this number x!

Input
The first line contains three integers n, a and b (1 ≤ a, b ≤ 100, 2 ≤ n ≤ a + b) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.

Output
Print the maximum possible number x such that Ivan can distribute the cake in such a way that each plate will contain at least x pieces of cake.

Examples
Input
5 2 3
Output
1
Input
4 7 10
Output
3
Note
In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.

In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.

题意： 两块蛋糕分n份，令最小的一块最大
思路： 直接枚举分最小的即可

#include
#include
#include
#include
using namespace std;
const int maxn = 2e5 + 7;

int main()
{
    int n,a,b;
    scanf("%d%d%d",&n,&a,&b);
    int ans = 0;
    if(a > b)
        swap(a,b);
    for(int i = 1;i <= a;i++)
    {
        if(a / i + b / i >= n)
            ans = max(i,ans);
    }
    printf("%d\n",ans);
}