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Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:

• each cut should be straight (horizontal or vertical);
• each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
• each cut should go inside the whole chocolate bar, and all cuts must be distinct.

The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.

Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.

Input

A single line contains three integers n, m, k (1 ≤ n, m ≤ 109; 1 ≤ k ≤ 2·109).

Output

Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.

题意:给一个n*m大小的方格网，再给一个切割次数k 问切割后最小的方格最大是多少。 

思路： 这题去想一个公式比较好，n 个格子被切割k1 次后 最小的方格是多大。可以知道是n /（k+1）所以就有最小的方格面积是    n/（k1+1）*m/(k2+1)即 n*m/(k1+1)*(k2+1)  就是求 看

（k1+1）*(k2+1)最小是多少。因为k1+k2=k,由于基本不等式k1和k2 相差大的时候就是可以的。不过要保证k1+1<=n,k2+1<=m

#include<iostream>#include<algorithm>using namespace std;int main(){	int n,m,k;	while(cin>>n>>m>>k){	if(k+2>n+m)	{		cout<<-1<<endl;		continue;	}	if(k+2==n+m)	{		cout<<1<<endl;		continue;	}	int ans;	if(k+1<=n) ans=m*(n/(k+1));	else       ans=m/(k-n+2);	if(k+1>m) ans=max(ans,n/(k-m+2));	else      ans=max(ans,n*(m/(k+1)));	cout<<ans<<endl;	}}