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!
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.
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.
5 2 3
1
4 7 10
3
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.
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> using namespace std; int main() { int n,a,b,m = 0; cin>>n>>a>>b; for(int i = 1;i < n;i ++) { int d = min(a/i,b/(n - i)); if(d > m)m = d; } cout<<m; }
