# Codeforces 450 C. Jzzhu and Chocolate, codeforcesjzzhu

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Codeforces 450 C. Jzzhu and Chocolate, codeforcesjzzhu

Quadratic functions n * m/(x + 1) * (k-x + 1) are quadratic functions with symmetric axes> 0 opening down, directly consider x = 0 | x = n-1 | x = M-1 and so on .....

C. Jzzhu and Chocolatetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Jzzhu has a big rectangular chocolate bar that consistsNLimit × limitMUnit squares. He wants to cut this bar exactlyKTimes. Each cut must meet the following requirements:

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

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

Imagine Jzzhu have madeKCuts 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 exactlyKCuts? The area of a chocolate piece is the number of unit squares in it.

Input

A single line contains three integersN, Bytes,M, Bytes,K(1 digit ≤ DigitN, Bytes,MLimit ≤ limit 109; 1 limit ≤ limitKLimit ≤ limit 2-109 ).

Output

Output a single integer representing the answer. If it is impossible to cut the big chocolateKTimes, 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 rows × every 3 chocolate 4 times.

`#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long int LL;LL n,m,k,ans=0LL;int main(){    cin>>n>>m>>k;    if(n+m-2<k)    {        puts("-1"); return 0;    }    if(n+m-2==k)    {        puts("1"); return 0;    }    LL t;    if(m>=k+1)        ans=max(ans,n*(m/(k+1)));    else    {        t=k-m+1;        ans=max(ans,n/(t+1));    }    if(n>=k+1)        ans=max(ans,m*(n/(k+1)));    else    {        t=k-n+1;        ans=max(ans,m/(t+1));    }    cout<<ans<<endl;    return 0;}`

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