Codeforces Round #257 (Div. 2) C. Jzzhu and Chocolate (数学)

C. Jzzhu and Chocolate

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 for5 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 chocolatek times, print -1.

Sample test(s)

Input

3 4 1

Output

6

Input

6 4 2

Output

8

Input

2 3 4

Output

-1

Note

In the first sample, Jzzhu can cut the chocolate following the picture below:

In the second sample the optimal division looks like this:

In the third sample, it's impossible to cut a 2 × 3 chocolate4 times.

题意：一个n*m的巧克力，已经分好块了，让你切k刀，求能切最小的方案中的最大块含有的巧克力

思路：刚开始是想横竖平均分着切，但是画了一下发现不行，后来想到方案要最小，所以说切法就不一样了，不可能切的条件很容易想到，就是n-1+m-1<k，当能切的时候，优先切最长的，如果最长和最短都能全部切完k刀，那么，比较取最大值，如果不能，那就先从最长的开始切，然后再在短的切剩下的几刀，这样答案就出来了

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int main()
{
	ll n,m,k;
	while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)
	{
		ll ans;
		if(n-1+m-1<k)
		printf("-1\n");
		else
		{
			ans=-1;
			int bz=0;
			if(n-1>=k)
			bz=1,ans=max(ans,(n-n%(k+1))/(k+1)*m);
			if(m-1>=k)
			bz=1,ans=max(ans,(m-m%(k+1))/(k+1)*n);
			if(bz==0)
			{
				if(n>m)
				ans=max(ans,(m-m%(k-n+2))/(k-n+2));
				else
				ans=max(ans,(n-n%(k-m+2))/(k-m+2));
			}
			printf("%I64d\n",ans);
		}
		
	}
	return 0;
}