【19.05%】【codeforces 680D】Bear and Tower of Cubes
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Limak is a little polar bear. He plays by building towers from blocks. Every block is a cube with positive integer length of side. Limak has infinitely many blocks of each side length.
A block with side a has volume a3. A tower consisting of blocks with sides a1, a2, …, ak has the total volume a13 + a23 + … + ak3.
Limak is going to build a tower. First, he asks you to tell him a positive integer X — the required total volume of the tower. Then, Limak adds new blocks greedily, one by one. Each time he adds the biggest block such that the total volume doesn’t exceed X.
Limak asks you to choose X not greater than m. Also, he wants to maximize the number of blocks in the tower at the end (however, he still behaves greedily). Secondarily, he wants to maximize X.
Can you help Limak? Find the maximum number of blocks his tower can have and the maximum X ≤ m that results this number of blocks.
Input
The only line of the input contains one integer m (1 ≤ m ≤ 1015), meaning that Limak wants you to choose X between 1 and m, inclusive.
Output
Print two integers — the maximum number of blocks in the tower and the maximum required total volume X, resulting in the maximum number of blocks.
Examples
input
48
output
9 42
input
6
output
6 6
Note
In the first sample test, there will be 9 blocks if you choose X = 23 or X = 42. Limak wants to maximize X secondarily so you should choose 42.
In more detail, after choosing X = 42 the process of building a tower is:
Limak takes a block with side 3 because it’s the biggest block with volume not greater than 42. The remaining volume is 42 - 27 = 15.
The second added block has side 2, so the remaining volume is 15 - 8 = 7.
Finally, Limak adds 7 blocks with side 1, one by one.
So, there are 9 blocks in the tower. The total volume is is 33 + 23 + 7·13 = 27 + 8 + 7 = 42.
【题解】
递归求解;
一层层地减少问题的规模;
类似分治的思想;
dfs(int w);表示m=w的时候问题的解是什么;
一开始的时候调用dfs(m);
然后,对于”当前”的w(也即这个子问题中的m);
我们有两种决策
1.直接取最大的;
则我们找到最大的p;
满足p^3<=w;
然后体积递增p^3,方块数递增1,递归求解dfs(w-p^3);
也即m变成了w-p^3,体积可以是1..w-p^3中的任意的最优解;
2.不直接取最大的;换成取中间的某个数字;
对p考虑;
这里的p仍然是上面的p;
我们可以递归求解dfs(p^3-1);
这样,下一层递归里面选最大的就会变成选(p-1)^3;
再往下递归一层就是dfs(p^3-1 - (p-1)^3);
而当我们递归dfs(p^3-1 - (p-1)^3)的时候,方块的数目也只是1;
而dfs(p^3-1 - (p-1)^3)不一定就比dfs(w-p^3)(调用时方块数也是1)大,则我们先比较一下这两个值的大小。如果前者大于后者。则说明前者有可能弄出来更优的解。否则的话就没必要递归dfs(p^3-1);
那有没有可能调用递归dfs((p-1)^3-1)呢?答案是否定的;
因为
f(p) = p^3 - 1 - (p-1)^3在0到正无穷上是一个单调递增的函数;
则dfs((p-1)^3-(p-2)^3)肯定不能弄出比dfs(p^3-1-(p-1)^3)更优的答案;
快速找出那个p可以用二分;
还不理解就多想想吧。
这题想了挺久的。上面也只是我自己的理解。
#include <cstdio>
#include <map>
#include <vector>
#define LL long long
using namespace std;
LL m;
pair <LL, LL> ans;
map <LL, pair<LL, LL> > fre;
LL get_max(LL x)
{
LL l = 1, r = (1e5 + 10);
LL ans;
while (l <= r)
{
LL mid = (l + r) >> 1;
LL t = mid*mid*mid;
if (t <= x)
{
ans = t;
l = mid + 1;
}
else
r = mid - 1;
}
return ans;
}
pair <LL, LL> dfs(LL x)
{
if (x == 0)
return make_pair(0, 0);
if (fre.count(x))
return fre[x];
pair <LL, LL> &temp = fre[x];
LL p3 = get_max(x);
temp.first = 0;
temp = dfs(x - p3);
temp.first++;
temp.second += p3;
pair <LL, LL> temp2;
LL t2 = 0;
if (p3 - 1 > 0)
{
t2 = p3 - 1 - get_max(p3 - 1);
}
if (t2 > (x - p3))
temp2 = dfs(p3 - 1);
else
temp2.first = 0;
if (temp2.first > temp.first)
temp = temp2;
return temp;
}
int main()
{
//freopen("F:\\rush.txt", "r", stdin);
scanf("%I64d", &m);
ans = dfs(m);
printf("%I64d %I64d\n", ans.first, ans.second);
return 0;
}
【19.05%】【codeforces 680D】Bear and Tower of Cubes的更多相关文章
- codeforces 680D D. Bear and Tower of Cubes(dfs+贪心)
题目链接: D. Bear and Tower of Cubes time limit per test 2 seconds memory limit per test 256 megabytes i ...
- 【CodeForces】679 B. Bear and Tower of Cubes
[题目]B. Bear and Tower of Cubes [题意]有若干积木体积为1^3,2^3,...k^3,对于一个总体积X要求每次贪心地取<=X的最大积木拼上去(每个只能取一次)最后总 ...
- Codeforces 680D Bear and Tower of Cubes 贪心 DFS
链接 Codeforces 680D Bear and Tower of Cubes 题意 求一个不超过 \(m\) 的最大体积 \(X\), 每次选一个最大的 \(x\) 使得 \(x^3\) 不超 ...
- Codeforces 680D - Bear and Tower of Cubes
680D - Bear and Tower of Cubes 思路:dfs+贪心,设剩余的体积为res,存在a,使得a3 ≤ res,每次取边长为a的立方体或者边长为a-1的立方体(这时体积上限变成a ...
- Codeforces Round #356 (Div. 2) D. Bear and Tower of Cubes dfs
D. Bear and Tower of Cubes 题目连接: http://www.codeforces.com/contest/680/problem/D Description Limak i ...
- 【19.05%】【codeforces 731F】 Video Cards
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- 【2018.05.11 智能驾驶/汽车电子】非技术向:关于Simulink和AutoSar的几种观点
最近看到几篇关于Simulink及AutoSar的Blog和Paper,感觉比较有意思,转载备忘之. 1. 看衰Simulink及AutoSar From:Tumiz的技术天地 https://blo ...
- 【2018.05.10 智能驾驶/汽车电子】AutoSar Database-ARXML及Vector Database-DBC的对比
最近使用python-canmatrix对can通信矩阵进行编辑转换时,发现arxml可以很容易转换为dbc,而dbc转arxml却需要费一番周折,需要额外处理添加一些信息. 注意:这里存疑,还是需要 ...
- 【 BowWow and the Timetable CodeForces - 1204A 】【思维】
题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...
随机推荐
- Docker初步了解 2016-10-30 20:46 279人阅读 评论(31) 收藏
什么是docker? Docker是一个开源的引擎,可以轻松的为任何应用创建一个轻量级的.可移植的.自给自足的容器. Docker本质上是一种软件,让用户创建镜像(很像虚拟机的模板),并且随后在容器里 ...
- MaxCompute客户端(odpscmd)在windows命令行下查询中文乱码问题处理实践
MaxCompute客户端工具是阿里云大数据计算服务MaxCompue产品官方客户端工具,通过客户端工具可以连接MaxCompute项目,完成包括数据管理.数据上下传.作业执行.用户及授权管理等各项操 ...
- React Native等比放大不丢失图片
9月11号 0.33版本,resizeMode中添加了center, 可以实现一样的功能.不过希望这篇文章还能帮助大家. 之前我们学习了从零学React Native之08Image组件 大家可以发现 ...
- 伪静态的实现方法:IIS环境下配置
URL 静态化可以提高搜索引擎抓取,开启本功能需要对 Web 服务器增加相应的 Rewrite 规则,且会轻微增加服务器负担.本教程讲解如何在 IIS 环境下配置各个产品的 Rewrite 规则. 下 ...
- PHP 7.0新增特性详解
https://www.cnblogs.com/riverdubu/archive/2017/03/22/6434705.html 开始介绍PHP7.0新特性,具体的可以参照官网的介绍,我来挑一些给大 ...
- 30 Cool Open Source Software I Discovered in 2013
30 Cool Open Source Software I Discovered in 2013 #1 Replicant – Fully free Android distribution Rep ...
- python selenium 测试浏览器(IE,FF,Chrome)
browser_engine.py # coding=utf-8 from selenium import webdriver class BrowserEngine(object): "& ...
- JQuery---高级选择器
一.派生选择器 例如:$('#bavBar a') 二.孩子选择器 例如:$('body > p') 三.相邻兄弟选择器 例如:$('h2 + div') 四.属性选择器 1.$('img[a ...
- 系统学习前端之FormData详解
FormData 1. 概述 FormData类型其实是在XMLHttpRequest 2级定义的,它是为序列化表以及创建与表单格式相同的数据(当然是用于XHR传输)提供便利. 2. 构造函数 创建一 ...
- 浮动,定位,flex布局
什么是文档流 英文原文是:Normal flow. In CSS 2.1, normal flow includes block formatting of block-level boxes, in ...