题目链接

题意：



题解：











AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9+7;
//const int mod = 998244353;
const double eps = 1e-10;
const double pi = acos(-1.0);
const int maxn = 1e6+10;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
double dfs(double x, double y, int n) {
    if (n == 1)
        return max(x, y) / min(x, y);
    double a = x / n, b = y / n, ans = inf;
    for (int i = 1; i <= n / 2; i++)
        ans = min(ans, min(max(dfs(i * a, y, i), dfs(x - i * a, y, n - i)), max(dfs(x, i * b, i), dfs(x, y - i * b, n - i))));
    return ans;
}

int main()
{
    int x, y, n;
    scanf("%d%d%d", &x, &y, &n);
    printf("%.6f\n", dfs(x, y, n));
    return 0;
}