codeforces 1140D(區間dp/思惟題)

D. Minimum Triangulation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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

input

Copy

3

output

Copy

6

input

Copy

4

output

Copy

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.

 題意：給你一個有n個頂點的正多邊形，頂點編號從1-n，如今你要把這個正多邊形劃分爲三角形，並且每一個三角形的面積都不相交，三角形的花費定義爲三角形頂點編號的乘積，問花費的最少代價。

 

思路：簡單的區間dp，對於dp[i][j]的求解，咱們只要以i，j爲底邊，枚舉頂點k（i<k<j），劃分出三角形（i，j，k），而後咱們就將要求的值分爲dp[i][k]+dp[k][j]+三角形（i，j，k）的花費，找到花費最小的值就能夠了。

若是敏感的話應該能夠發現對於正多邊形的劃分，就是劃分爲（1，2 ，3），（1，3，4），（1，4，5）。。。。，（1，n-1，n）時它的花費是最小的，那麼最後的答案就是2*3+3*4+4*5+.......+（n-1）*n。

 

#include<cstdio> #include<algorithm>
using namespace std; const int INF=1e9; int dp[510][510]; int main(){ int n; scanf("%d",&n); for(int j=2;j<=n-1;j++){ for(int i=1;i+j<=n;i++){ dp[i][i+j]=INF;//找最小值，先初始化爲無窮大 
            for(int k=i+1;k<i+j;k++){ dp[i][i+j]=min(dp[i][i+j],dp[i][k]+dp[k][i+j]+i*(i+j)*k); } } } printf("%d\n",dp[1][n]); return 0; } 

View Code

#include<cstdio> #include<algorithm>
using namespace std; const int INF=1e9; int main(){ int n; scanf("%d",&n); int ans=0; for(int i=3;i<=n;i++) ans+=i*(i-1); printf("%d\n",ans); return 0; }

View Code