A12560 | Minimum Triangulation
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内存限制256MB
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题目描述
You are given a regular polygon with $n$ vertices labeled from $1$ to $n$ in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.
Calculate the minimum weight among all triangulations of the polygon.
Calculate the minimum weight among all triangulations of the polygon.
输入格式
The first line contains single integer $n$ ( $3 \le n \le 500$ ) — the number of vertices in the regular polygon.
输出格式
Print one integer — the minimum weight among all triangulations of the given polygon.
输入输出样例
输入 #1
3
输出 #1
6
输入 #2
4
输出 #2
18
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) $P$ into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is $P$ .
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is $1 \cdot 2 \cdot 3 = 6$ .
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal $1-3$ so answer is $1 \cdot 2 \cdot 3 + 1 \cdot 3 \cdot 4 = 6 + 12 = 18$ .
In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is $1 \cdot 2 \cdot 3 = 6$ .
In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal $1-3$ so answer is $1 \cdot 2 \cdot 3 + 1 \cdot 3 \cdot 4 = 6 + 12 = 18$ .
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评测结果:Accepted