POJ 1106

先判断是否在圆内，然后用叉积判断是否在180度内。枚举判断就可以了。。。

感觉是数据弱了。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const double eps=0.00000001;
struct point{
	double x,y;
}p[1050],circle;
double rad;
int n,ans;

double dist(double x1,double y1, double x2,double y2){
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));	
}

bool multi(point a,point b){
	if((a.x-circle.x)*(b.y-circle.y)-(a.y-circle.y)*(b.x-circle.x)>=0)
	return true;
	return false;
}

void slove(){
	int tmp;
	for(int i=0;i<n;i++){
		tmp=0;
		for(int j=0;j<n;j++){
			if(multi(p[i],p[j])){
				tmp++;
			}
		}
		if(tmp>ans) ans=tmp;
	}
}

int main(){
	int ni; double x,y;
	while(scanf("%lf%lf%lf",&circle.x,&circle.y,&rad)!=EOF){
		if(rad<0) break;
		scanf("%d",&ni); n=0;
		for(int i=0;i<ni;i++){
			scanf("%lf%lf",&x,&y);
			if(dist(x,y,circle.x,circle.y)<=rad){
				p[n].x=x; p[n].y=y;
				n++;
			}
		}
	//	cout<<n<<endl;
		if(rad==0) { printf("0
"); continue; }
		ans=0;
		slove();
		printf("%d
",ans);
	}
	return 0;
}