Codeforces Beta Round 51
E
Very simple problem
Codeforces
2500
原题
← D→
翻译中…

You are given a convex polygon. Count, please, the number of triangles that contain a given point in the plane and their vertices are the vertices of the polygon. It is guaranteed, that the point doesn't lie on the sides and the diagonals of the polygon.

翻译中…

Input

翻译中…

The first line contains integer $*n*$ — the number of vertices of the polygon ($3 ≤ *n* ≤ 100000$). The polygon description is following: $*n*$ lines containing coordinates of the vertices in clockwise order (integer $*x*$ and $*y*$ not greater than $10^{9}$ by absolute value). It is guaranteed that the given polygon is nondegenerate and convex (no three points lie on the same line).

翻译中…

The next line contains integer $*t*$ ($1 ≤ *t* ≤ 20$) — the number of points which you should count the answer for. It is followed by $*t*$ lines with coordinates of the points (integer $*x*$ and $*y*$ not greater than $10^{9}$ by absolute value).

翻译中…

Output

翻译中…

The output should contain $*t*$ integer numbers, each on a separate line, where $*i*$-th number is the answer for the $*i*$-th point.

翻译中…

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

翻译中…

样例

Input #1
4
5 0
0 0
0 5
5 5
1
1 3
Output #1
2
InputCopy OutputCopy
Input #2
3
0 0
0 5
5 0
2
1 1
10 10
Output #2
1
0
InputCopy OutputCopy
Input #3
5
7 6
6 3
4 1
1 2
2 4
4
3 3
2 3
5 5
4 2
Output #3
5
3
3
4
InputCopy OutputCopy
题目信息
题号
55E
时间限制
3 s
内存限制
256 MB
标签
计算几何
双指针

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。