CodeForces 449A - Jzzhu and Chocolate(贪心)

A. 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 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.

Examples

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 chocolate 4 times.

/*
题目链接：http://codeforces.com/contest/450/problem/C

题目意思：给出一个 n * m 大小的chocolate bar，你需要在这个bar上切 k 刀，使得最小的部分面积尽可能大，
求出这个被划分后的最小部分面积最大可以为多少。如果这个chocolate bar 不能切成 k 部分，则输出-1。
注意，每一刀需要符合3个条件：1、打横切或者打竖切； 2、每一刀只能经过unit square（即1*1的单元bar）的边，
也就是说不能把一个单元bar损坏，要完整；
3、每一刀只能在整个chocolate bar的里面操作，
也就是说，外围的四条边是不允许切的。
还有一个条件就是，每一刀都是不相同的。

if (n-1 >= k || m-1 >= k) 全部都横切或者竖切
else  先把横(或竖)全切了  把剩下的倒数拿去切竖(横)的
因为要求最小块 所以直接 除就行了
*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <stack>
#include <string>
#include <sstream>
#include <map>
#include <set>
#define pi acos(-1.0)
#define LL long long
#define ULL unsigned long long
#define inf 0x3f3f3f3f
#define INF 1e18#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
typedef pair<int, int> P;
const double eps = 1e-10;
const int maxn = 1e6 + 5;
const int N = 1e4 + 5;
const int mod = 1e8;

int main(void)
{
//	freopen("in.txt","r", stdin);
LL n, m, k;
// 假定n为竖的  m为横的
while (cin >> n >> m >> k)
{
if (n + m - 2 < k){
cout << "-1" << endl;
continue;
}
LL ans = 0;
if (n-1 >= k)	// 横切k次 因为切了k次有会有k+1份 所以n/(k+1)
ans = max(ans, m * (n/(k+1)));
if (m-1 >= k)	// 竖切k次
ans = max(ans, n * (m/(k+1)));
if (n-1 < k)	// 横切n-1次 竖切k-(n-1)次
ans = max(ans, m / (k-(n-1)+1));
if (m-1 < k)	// 竖切m-1次 横切k-(m-1)次
ans = max(ans, n / (k-(m-1)+1));
cout << ans << endl;
}

return 0;
}