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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，現在你要把這個正多邊形劃分為三角形，而且每個三角形的面積都不相交，三角形的花費定義為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，我們只要以i，j為底邊，枚舉頂點k（i<k<j），劃分出三角形（i，j，k），然後我們就將要求的值分為dp[i][k]+dp[k][j]+三角形（i，j，k）的花費，找到花費最小的值就可以了。

如果敏感的話應該可以發現對於正多邊形的劃分，就是劃分為（1，2 ，3），（1，3，4），（1，4，5）。。。。，（1，n-1，n）時它的花費是最小的，那麽最後的答案就是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，先初始化為無窮大 
            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

codeforces 1140D(區間dp/思維題)