CodeForces - 821B Okabe and Banana Trees 解题

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　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 + y bananas. 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.

 

 

这道题主要是找一个规律，x=m*(b-y)，知道x以后就可以直接遍历，用公式算出来以后就可以进行比较，最后把大的留下。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<cmath>
#include<cctype>
#include<ctime>

using namespace std;

int main()
{
    long long m,b,x,num=0;
    scanf("%lld%lld",&m,&b);
    for(int i=0;i<=b;i++)
    {
        long long sum=0;
        x=m*(b-i);
        sum=(x*(x+1)/2)*(i+1)+(i*(i+1)/2)*(x+1);
        if(sum>num) num=sum;
    }
    printf("%lld\n",num);
    return 0;
}