B. Okabe and Banana Treestime 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.
Examplesinput
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.
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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;
}
