Codeforces Round #257 (Div. 2) C. Jzzhu and Chocolate-白红宇

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

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发布时间：2019-06-24

本文共 2310 字，大约阅读时间需要 7 分钟。

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.

$\text{[math]}$

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.

Sample test(s)

Input

3 4 1

Output

Input

6 4 2

Output

Input

2 3 4

Output

-1

Note

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

$\text{[math]}$

In the second sample the optimal division looks like this:

$\text{[math]}$

In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.

1 #include
          
            2 #include
           
             3 #include
            
              4 #include
             
               5 #include
              
                6 #include
               
                 7 #include
                
                  8 #include
                 
                   9 #include
                  
                   10 #include
                   
                    11 #include
                    12 #include
                     
                      13 #include
                      
                       14 using namespace std;15 #define M 100000000716 __int64 n,m;17 __int64 f(__int64 a, __int64 b)18 {19 if(a<0 || b<0) return 0;20 return max((n/(a+1))*(m/(b+1)), (n/(b+1))*(m/(a+1)));21 }22 23 int main()24 {25 __int64 k,ans=0;26 27 scanf("%I64d%I64d%I64d",&n, &m, &k);28 29 if(k>n+m-2) printf("-1\n");30 else if(k==n+m-2) printf("1\n");31 else32 {33 ans=max(ans, f(0, k));34 ans=max(ans, f(m-1, k-m+1));35 ans=max(ans, f(n-1, k-n+1));36 printf("%I64d\n", ans);37 }38 39 return 0;40 }

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