DAY15.2 D. Minimum Triangulation

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a regular polygon with nn vertices labeled from 11 to nn 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.

Input
The first line contains single integer nn (3≤n≤5003≤n≤500) — the number of vertices in the regular polygon.

Output
Print one integer — the minimum weight among all triangulations of the given polygon.

Examples
inputCopy
3
outputCopy
6
inputCopy
4
outputCopy
18
Note
According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) PP into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is PP.

In the first example the polygon is a triangle, so we don’t need to cut it further, so the answer is 1⋅2⋅3=61⋅2⋅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 1−31−3 so answer is 1⋅2⋅3+1⋅3⋅4=6+12=181⋅2⋅3+1⋅3⋅4=6+12=18.

思路&过程

要想最小，就都和1连。

代码：

#include<iostream>
using namespace std;

int main()
{
	long long n,i,j,ans;
	cin>>n;
	ans=0;
	while(n>=3)
	{
		ans=ans+n*(n-1);
		n--;
	}
	cout<<ans<<endl;
	return 0;
 }