对多边形的每个边计算和原点的夹角范围，求并得到总夹角范围 alpha
答案就是 min(alpha, 2 * pi) * k * h
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-11 20:01:23
File Name: pku1031.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{ for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) { for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
const double eps = 1e-10;
const double pi = acos(-1.0);
const int maxn = 110;
typedef struct point_t
{
double x, y;
};
typedef struct interval_t
{
double l, r;
};
double get_angle(double x, double y)
{
if (abs(x) < eps)
{
if (y > 0) return pi / 2.0;
else return 3.0 * pi / 2.0;
}
else
{
double t = atan(y / x);
if (x < 0)
t += pi;
if (t < 0) t += 2.0 * pi;
return t;
}
}
interval_t get_range(point_t a, point_t b)
{
if (a.x * b.y - a.y * b.x < eps)
{
swap(a, b);
}
double t1 = get_angle(a.x, a.y);
double t2 = get_angle(b.x, b.y);
if (t2 < t1 - eps) t2 += 2.0 * pi;
interval_t ret;
ret.l = t1;
ret.r = t2;
return ret;
}
bool operator <(const interval_t &a, const interval_t &b)
{
return a.l < b.l || a.l == b.l && a.r < b.r;
}
bool between(const double &a, const double &x, const double &b)
{
return x > a - eps && x < b + eps;
}
point_t p[maxn];
interval_t interval[maxn];
int main()
{
double k, h;
int n;
scanf("%lf%lf%d", &k, &h, &n);
{
for (int i = 0; i < n; i++)
scanf("%lf%lf", &p[i].x, &p[i].y);
for (int i = 0; i < n; i++)
interval[i] = get_range(p[i], p[(i + 1) % n]);
double l = interval[0].l, r = interval[0].r;
for (int i = 1; i < n; i++)
{
for (int j = -1; j <= 1; j++)
{
if (between(l, interval[i].l + j * 2 * pi, r) || between(l, interval[i].r + j * 2 * pi, r))
{
l = min(l, interval[i].l + j * 2 * pi);
r = max(interval[i].r + j * 2 * pi, r);
break;
}
}
}
printf("%.2lf\n", min(r - l, 2.0 * pi) * k * h);
}
return 0;
}