Codeforces 821B-Okabe and Banana Trees

最新推荐文章于 2017-10-09 08:42:15 发布

原创最新推荐文章于 2017-10-09 08:42:15 发布 · 478 阅读

0 ·

CC 4.0 BY-SA版权

----水题同时被 2 个专栏收录

234 篇文章

订阅专栏

Codeforces

140 篇文章

订阅专栏

本题中，Okabe为了实验需要收集香蕉，决定在森林中切割香蕉树。题目要求在一条给定直线之下画出一个轴对齐的矩形区域，以收集该区域内所有点(x, y)处的香蕉(x + y)，目标是找到能够收集最多香蕉的矩形位置。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 $\text{[math]}$ . 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

$\text{[math]}$

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

题意：给你一条直线，在第一象限且在直线下画一个矩形，每个点（x，y）的香蕉数为x+y，矩形内包括边界上的香蕉可采，问最大香蕉数

解题思路：按y讨论，香蕉数可用y计算得到，首先计算出最右边界，然后用等差数列求和导出香蕉总数公式，若不这样计算，会超时

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#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, b, sum, ans;
	while (~scanf("%lld %lld", &m, &b))
	{
		ans = 0;
		for (LL i = b; i >= 0; i--)
		{
			LL k = b*m - i*m;
			sum = k*(k + 1)*(i + 1) / 2 + i*(i + 1)*(k + 1) / 2;
			ans = max(ans, sum);
		}
		printf("%lld\n", ans);
	}
	return 0;
}