CF437E.The Child and Polygon
普及/提高-
通过率:0%
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题目描述
This time our child has a simple polygon. He has to find the number of ways to split the polygon into non-degenerate triangles, each way must satisfy the following requirements:
- each vertex of each triangle is one of the polygon vertex;
- each side of the polygon must be the side of exactly one triangle;
- the area of intersection of every two triangles equals to zero, and the sum of all areas of triangles equals to the area of the polygon;
- each triangle must be completely inside the polygon;
- each side of each triangle must contain exactly two vertices of the polygon.
The picture below depicts an example of a correct splitting.
Please, help the child. Calculate the described number of ways modulo 1000000007 (109+7) for him.
输入格式
The first line contains one integer n (3<=n<=200) — the number of vertices of the polygon. Then follow n lines, each line containing two integers. The i -th line contains xi,yi (∣xi∣,∣yi∣<=107) — the i -th vertex of the polygon in clockwise or counterclockwise order.
It's guaranteed that the polygon is simple.
输出格式
Output the number of ways modulo 1000000007 (109+7) .
输入输出样例
输入#1
4 0 0 0 1 1 1 1 0
输出#1
2
输入#2
4 0 0 1 0 0 1 -1 0
输出#2
1
输入#3
5 0 0 1 0 1 1 0 1 -2 -1
输出#3
3
说明/提示
In the first sample, there are two possible splittings:
In the second sample, there are only one possible splitting: