Codeforces Round #257 (Div. 2)C 貪心

標籤：acm   codeforces   

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

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.

題解

好逗的題目。就是一個網格狀的物品，n行m列，筆直直的切k刀，問所有的可能性中，最大的，每次切出來的結果最小的塊的大小。

自己瞎貪心跪了。賽後發現其實可以枚舉可能的在行這個方向切得刀數r，計算資料行方向切的刀數o，然後更新答案的方法。當然，枚舉的時候也是只枚舉因數，這樣就可以把複雜度從O(N)降到O( sqrt(N) )。中間在加上一些最佳化，然後就沒事了。

程式碼範例

/*******************************************************************************       COPYRIGHT NOTICE*       Copyright (c) 2014 All rights reserved*       ----Stay Hungry Stay Foolish----** @author:Shen* @name         :C* @file         :G:\My Source Code\【ACM】比賽\0719 - CF\C.cpp* @date         :2014/07/19 20:57* @algorithm    :Greedy******************************************************************************/#include <bits/stdc++.h>using namespace std;template<class T>inline bool updateMin(T& a, T b){ return a > b ? a = b, 1 : 0; }template<class T>inline bool updateMax(T& a, T b){ return a < b ? a = b, 1 : 0; }typedef long long int64;int64 n, m, k, ans;int64 run(){    if (n - 1 + m - 1 < k) ans = -1;for (int64 i = 1; i <= n; i++)    {int64 r = n / (n / i); i = r;if (k - (r - 1) > m - 1) continue;int64 o = max(k - (r - 1) + 1, 1LL);updateMax(ans, 1LL * (n / r) * (m / o));}    return ans;}int main(){    cin >> n >> m >> k;    cout << run() << endl;    return 0;}

Codeforces Round #257 (Div. 2)C 貪心