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 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.

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.

 #include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
#define M 1000000007
__int64 n,m;
__int64 f(__int64 a, __int64 b)
{
if(a< || b<) return ;
return max((n/(a+))*(m/(b+)), (n/(b+))*(m/(a+)));
} int main()
{
__int64 k,ans=; scanf("%I64d%I64d%I64d",&n, &m, &k); if(k>n+m-) printf("-1\n");
else if(k==n+m-) printf("1\n");
else
{
ans=max(ans, f(, k));
ans=max(ans, f(m-, k-m+));
ans=max(ans, f(n-, k-n+));
printf("%I64d\n", ans);
} return ;
}

Codeforces Round #257 (Div. 2) C. Jzzhu and Chocolate的更多相关文章

  1. Codeforces Round #257 (Div. 1)449A - Jzzhu and Chocolate(贪婪、数学)

    主题链接:http://codeforces.com/problemset/problem/449/A ------------------------------------------------ ...

  2. Codeforces Round #257 (Div. 2) A. Jzzhu and Children(简单题)

    题目链接:http://codeforces.com/problemset/problem/450/A ------------------------------------------------ ...

  3. Codeforces Round #257(Div. 2) B. Jzzhu and Sequences(矩阵高速幂)

    题目链接:http://codeforces.com/problemset/problem/450/B B. Jzzhu and Sequences time limit per test 1 sec ...

  4. Codeforces Round #257 (Div. 1) C. Jzzhu and Apples (素数筛)

    题目链接:http://codeforces.com/problemset/problem/449/C 给你n个数,从1到n.然后从这些数中挑选出不互质的数对最多有多少对. 先是素数筛,显然2的倍数的 ...

  5. Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences (矩阵快速幂)

    题目链接:http://codeforces.com/problemset/problem/450/B 题意很好懂,矩阵快速幂模版题. /* | 1, -1 | | fn | | 1, 0 | | f ...

  6. Codeforces Round #257 (Div. 2) B Jzzhu and Sequences

    Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, ple ...

  7. Codeforces Round #257 (Div. 1) D - Jzzhu and Numbers 容斥原理 + SOS dp

    D - Jzzhu and Numbers 这个容斥没想出来... 我好菜啊.. f[ S ] 表示若干个数 & 的值 & S == S得 方案数, 然后用这个去容斥. 求f[ S ] ...

  8. Codeforces Round #257 (Div. 2) A. Jzzhu and Children

    A. Jzzhu and Children time limit per test 1 second memory limit per test 256 megabytes input standar ...

  9. Codeforces Round #257(Div.2) D Jzzhu and Cities --SPFA

    题意:n个城市,中间有m条道路(双向),再给出k条铁路,铁路直接从点1到点v,现在要拆掉一些铁路,在保证不影响每个点的最短距离(距离1)不变的情况下,问最多能删除多少条铁路 分析:先求一次最短路,铁路 ...

随机推荐

  1. Docker基础速成(一)

    Docker基础速成(一) 给亲爱的写的docker基础速成,按照步骤操作,实践出真知,希望有提纲挈领之功效 1.docker简介 Docker 轻量级容器,如图,类似于一个个集装箱,把复杂或者零散的 ...

  2. MySQL分布式集群之MyCAT(一)简介【转】

    隔了好久,才想起来更新博客,最近倒腾的数据库从Oracle换成了MySQL,研究了一段时间,感觉社区版的MySQL在各个方面都逊色于Oracle,Oracle真的好方便!好了,不废话,这次准备记录一些 ...

  3. 大数据系列之分布式计算批处理引擎MapReduce实践

    关于MR的工作原理不做过多叙述,本文将对MapReduce的实例WordCount(单词计数程序)做实践,从而理解MapReduce的工作机制. WordCount: 1.应用场景,在大量文件中存储了 ...

  4. Windows版Oracle重建EM---备注

    前提条件添加环境变量 ORACLE_HOSTNAME=<主机名:如:DESKTOP-P6J1a>ORACLE_SID=orclORACLE_UNQNAME=orcl 执行删除命令 C:\U ...

  5. http://s22.app1105796624.qqopenapp.com/

    http://s22.app1105796624.qqopenapp.com/ http://121.43.114.69/xiyou/app/js/ac_tx.js http://hiyouba.co ...

  6. Flask:使用Eclipse+PyDev插件编辑基于package的项目

    Windows 10家庭中文版,Python 3.6.4,Flask 1.0.2,Eclipse Oxygen.1a Release (4.7.1a),PyDev 6.3.2 本文记录了 使用Ecli ...

  7. Token机制,防止web页面重复提交

    1.业务要求:页面的数据只能被点击提交一次 2.发生原因: 由于重复点击或者网络重发,或者nginx重发等情况会导致数据被重复提交 3.解决办法: 集群环境:采用token加redis(redis单线 ...

  8. 最近公共祖先(LCA)模板

    以下转自:https://www.cnblogs.com/JVxie/p/4854719.html 首先是最近公共祖先的概念(什么是最近公共祖先?): 在一棵没有环的树上,每个节点肯定有其父亲节点和祖 ...

  9. 多线程 or I/O复用select/epoll

    1:多线程模型适用于处理短连接,且连接的打开关闭非常频繁的情形,但不适合处理长连接.线程模型默认情况下,在Linux下每个线程会开8M的栈空间,在TCP长连接的情况下,以2000/分钟的请求为例,几乎 ...

  10. 20155225 2006-2007-2 《Java程序设计》第四周学习总结

    20155225 2006-2007-2 <Java程序设计>第四周学习总结 教材学习内容总结 对"是一种"语法测试几次之后,总结一句:满足"是一种" ...