题意：一只狗被拴在杆子上，从起点开始按直线依次跑到给出的点最后回到起点问绕杆子几圈，逆时针为正，顺时针为负，撞到杆子输出Ouch!。
解法：用叉积判断方向，用余弦定理求出以杆子为顶点的角，加和除以2π，最后的答案处理因为精度问题wa了一篇orz……
代码：
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;
const double eps = 1e-;
struct node
{
double x, y;
} point[];
double cross(node p1, node p2, node p3)//叉积
{
return (p2.x - p1.x) * (p3.y - p1.y) - (p3.x - p1.x) * (p2.y - p1.y);
}
double len(node p1, node p2)//两点长度的平方
{
return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y);
}
double cal(node p1, node p2, node p3)//余弦定理
{
double a = len(p2, p3), b = len(p1, p3), c = len(p1, p2);
return acos((b + c - a) / 2.0 / sqrt(b * c));
}
int line(node p1, node p2, node p3)//判断叉积为0时杆是否在两点中间
{
return p3.x >= min(p1.x, p2.x) && p3.x <= max(p1.x, p2.x) && p3.y >= min(p1.y, p2.y) && p3.y <= max(p1.y, p2.y);
}
int main()
{
int n;
while(~scanf("%d", &n) && n)
{
node pole;
scanf("%lf%lf", &pole.x, &pole.y);
for(int i = ; i < n; i++)
scanf("%lf%lf", &point[i].x, &point[i].y);
double degree = 0.0;
int ans = ;
for(int i = ; i < n - ; i++)
{
double flag = cross(pole, point[i], point[i + ]);
double res = cal(pole, point[i], point[i + ]);
if((fabs(flag) <= eps) && line(point[i], point[i + ], pole))
{
ans = ;
break;
}
else if(flag > eps)
degree += res;
else if(flag < -eps)
degree -= res;
else
{
if(line(point[i], point[i + ], pole))
{
ans = ;
break;
}
}
}
double flag = cross(pole, point[n - ], point[]);
double res = cal(pole, point[n - ], point[]);
if((fabs(flag) <= eps) && line(point[n - ], point[], pole))
ans = ;
else if(flag > eps)
degree += res;
else if(flag < -eps)
degree -= res;
else if(line(point[n - ], point[], pole))
ans = ;
if(ans)
{
printf("Ouch!\n");
continue;
}
degree = degree * 0.5 / acos(-1.0);
char x[];
sprintf(x, "%.0lf", degree);
sscanf(x, "%d", &ans);
if(ans > )
printf("+");
printf("%d\n", ans);
}
return ;
}