B. Okabe and Banana Trees
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.

Consider the point (x, y) in the 2D plane such that x and y are
integers and 0 ≤ x, y. There is a tree in such a point, and it has x + ybananas.
There are no trees nor bananas in other points. Now, Okabe draws a line with equation .
Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it
can be a line segment or even a point.

Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.

Okabe is sure that the answer does not exceed 1018.
You can trust him.

Input

The first line of input contains two space-separated integers m and b (1 ≤ m ≤ 1000, 1 ≤ b ≤ 10000).

Output

Print the maximum number of bananas Okabe can get from the trees he cuts.

Examples
input
1 5
output
30
input
2 3
output
25
Note

The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas.

——————————————————————————————————————

题目的意思是给出一条斜线,在第一象限的斜线(可以轴上)上取一个点,它与原点形

成一个矩阵 每个点的值为横纵坐标之和,问矩阵内(上)所有点之和最大多少

思路:每一行各点之和是等差数列,矩阵中上一列比下一列的每个数大1 数学统计即可

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits> using namespace std; #define LL long long
const int INF = 0x3f3f3f3f; int main()
{
LL m,n;
while(~scanf("%lld%lld",&m,&n))
{
LL mx=-1;
for(int i=0;i<=n;i++)
{
LL x=((n-i)*m*((n-i)*m+1)/2)*(i+1)+((n-i)*m+1)*(1+i)*i/2;
mx=max(x,mx);
}
printf("%lld\n",mx); }
return 0;
}

Codeforces821B Okabe and Banana Trees 2017-06-28 15:18 25人阅读 评论(0) 收藏的更多相关文章

  1. Codeforces821C Okabe and Boxes 2017-06-28 15:24 35人阅读 评论(0) 收藏

    C. Okabe and Boxes time limit per test 3 seconds memory limit per test 256 megabytes input standard ...

  2. Codeforces821A Okabe and Future Gadget Laboratory 2017-06-28 14:55 80人阅读 评论(0) 收藏

    A. Okabe and Future Gadget Laboratory time limit per test 2 seconds memory limit per test 256 megaby ...

  3. Hdu2841 Visible Trees 2017-06-27 22:13 24人阅读 评论(0) 收藏

    Visible Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) To ...

  4. 2014/11/06 Oracle触发器初步 2014-11-06 09:03 49人阅读 评论(0) 收藏

    触发器我就不多解释了,保证数据的完整性的神器,嗯..也是减少程序员工作托管给数据库操作的好帮手.就不讲一些大道理了.通俗点,我们对数据库的操作,无非就是增 删 改 查. 触发器就是在删,改,增的时候( ...

  5. The Managed Metadata Service or Connection is currently not available 分类: Sharepoint 2015-07-09 13:28 5人阅读 评论(0) 收藏

    Does the following error message looks familiar to you? (When you go to Site Actions –> Site Sett ...

  6. NPOI 通用导出数据到Excel 分类: C# Helper 2014-11-04 16:06 246人阅读 评论(0) 收藏

    应用场景: 在项目中,经常遇到将数据库数据导出到Excel,针对这种情况做了个程序封装.工作原理:利用NPOI将SQL语句查询出的DataTable数据导出到Excel,所见即所得. 程序界面:   ...

  7. jQuery插件treeview点击节点名称不展开、收缩节点 分类: JavaScript 2014-06-16 20:28 539人阅读 评论(0) 收藏

    修改jquery.treeview.js文件中的applyClasses方法(注释掉两行代码): 修改后的applyClasses方法如下: applyClasses: function(settin ...

  8. cloud theory is a failure? 分类: Cloud Computing 2013-12-26 06:52 269人阅读 评论(0) 收藏

    since LTE came out, with thin client cloud computing  and broadband communication clouding 不攻自破了.but ...

  9. A simple problem 分类: 哈希 HDU 2015-08-06 08:06 1人阅读 评论(0) 收藏

    A simple problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) To ...

随机推荐

  1. 5E - A + B Again

    There must be many A + B problems in our HDOJ , now a new one is coming. Give you two hexadecimal in ...

  2. 三种简单排序算法(java实现)

    一.冒泡排序 算法思想:遍历待排序的数组,每次遍历比较相邻的两个元素,如果他们的排列顺序错误就交换他们的位置,经过一趟排序后,最大的元素会浮置数组的末端.重复操                   作 ...

  3. Max Chunks To Make Sorted LT769

    Given an array arr that is a permutation of [0, 1, ..., arr.length - 1], we split the array into som ...

  4. 【UI测试】--规范性

  5. Tomcat优化详细1

    在Tomcat和应用程序进行了压力测试后,如果您对应用程序的性能结果不太满意,就可以采取一些性能调整措施了,当然了前提是应用程序没有问题,我们这里只讲Tomcat的调整.由于Tomcat的运行依赖于J ...

  6. HACK字体安装

    参考:https://github.com/source-foundry/Hack Linux的 下载最新版本的Hack. 从存档中提取文件(.zip). 将字体文件复制到系统字体文件夹(通常/usr ...

  7. Flex + .Net从本地选择一个图片上传到服务器

    <mx:TextInput id="TxtFileName" editable="false" width="200"/> &l ...

  8. Redis的基操

    redis:通常BOLEAN操作类型,操作成功返回1,操作失败返回0 通常如果往设置的key插入值,但是这个key不存在,redis则会创建 向redis里的某个key插入多个值时,值和值之间用空格隔 ...

  9. android c 读写文件

    1.包含头文件 #include<unistd.h>#include<sys/types.h>#include<sys/stat.h>#include<fcn ...

  10. 查看Android应用所需权限(uses-permission)

    http://www.tuicool.com/articles/zq2meq MainActivity如下: package cc.testusespermission; import android ...