我心态崩了。
吃完饭回来，『哇这个圆与矩形交到底怎么求啊？？？』，顿了一秒，
嗯？？这不是傻逼板子题吗？
然后粘上了我圆和三角形交的板子。1A。
mxy到底有什么用啊？
这他妈这么傻逼的板子题。。我为什么四个小时都没意识到这是板子题。
这种题还能wa？这怎么wa啊？？
#include <bits/stdc++.h>
#define mp make_pair
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef double db;
const int maxn = 4e5+5;
const db eps=1e-6;
const db pi=acos(-1);
int sign(db k){
if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内
struct point{
db x,y;
point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
point operator * (db k1) const{return (point){x*k1,y*k1};}
point operator / (db k1) const{return (point){x/k1,y/k1};}
int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
// 逆时针旋转
point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
point turn90(){return (point){-y,x};}
bool operator < (const point k1) const{
int a=cmp(x,k1.x);
if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
}
db abs(){return sqrt(x*x+y*y);}
db abs2(){return x*x+y*y;}
db dis(point k1){return ((*this)-k1).abs();}
point unit(){db w=abs(); return (point){x/w,y/w};}
void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;}
void print(){printf("%.11lf %.11lf\n",x,y);}
db getw(){return atan2(y,x);}
point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}
int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)==-1);}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));}
// -pi -> pi
int compareangle (point k1,point k2){//极角排序+
return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>0);
}
point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影
point k=k2-k1;return k1+k*(dot(q-k1,k)/k.abs2());
}
point reflect(point k1,point k2,point q){return proj(k1,k2,q)*2-q;}
int clockwise(point k1,point k2,point k3){// k1 k2 k3 逆时针 1 顺时针 -1 否则 0
return sign(cross(k2-k1,k3-k1));
}
int checkLL(point k1,point k2,point k3,point k4){// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点
return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=0;
}
point getLL(point k1,point k2,point k3,point k4){
db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}
int intersect(db l1,db r1,db l2,db r2){
if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;
}
int checkSS(point k1,point k2,point k3,point k4){
return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&
sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
}
db disSP(point k1,point k2,point q){
point k3=proj(k1,k2,q);
if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));
}
db disSS(point k1,point k2,point k3,point k4){
if (checkSS(k1,k2,k3,k4)) return 0;
else return min(min(disSP(k1,k2,k3),disSP(k1,k2,k4)),min(disSP(k3,k4,k1),disSP(k3,k4,k2)));
}
int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;}
struct circle{
point o; db r;
void scan(){o.scan(); scanf("%lf",&r);}
int inside(point k){return cmp(r,o.dis(k));}
};
struct line{
// p[0]->p[1]
point p[2];
line(point k1,point k2){p[0]=k1; p[1]=k2;}
line(){}
point& operator [] (int k){return p[k];}
int include(point k){return sign(cross(p[1]-p[0],k-p[0]))>0;}
point dir(){return p[1]-p[0];}
line push(){ // 向外 ( 左手边 ) 平移 eps
const db eps = 1e-6;
point delta=(p[1]-p[0]).turn90().unit()*eps;
return {p[0]-delta,p[1]-delta};
}
};
point getLL(line k1,line k2){return getLL(k1[0],k1[1],k2[0],k2[1]);}
vector<point> getCL(circle k1,point k2,point k3){ // 沿着 k2->k3 方向给出 , 相切给出两个
point k=proj(k2,k3,k1.o); db d=k1.r*k1.r-(k-k1.o).abs2();
if (sign(d)==-1) return {};
point del=(k3-k2).unit()*sqrt(max((db)0.0,d)); return {k-del,k+del};
}
db getarea(circle k1,point k2,point k3){
// 圆 k1 与三角形 k2 k3 k1.o 的有向面积交
point k=k1.o; k1.o=k1.o-k; k2=k2-k; k3=k3-k;
int pd1=k1.inside(k2),pd2=k1.inside(k3);
vector<point>A=getCL(k1,k2,k3);
if (pd1>=0){
if (pd2>=0) return cross(k2,k3)/2;
return k1.r*k1.r*rad(A[1],k3)/2+cross(k2,A[1])/2;
} else if (pd2>=0){
return k1.r*k1.r*rad(k2,A[0])/2+cross(A[0],k3)/2;
}else{
int pd=cmp(k1.r,disSP(k2,k3,k1.o));
if (pd<=0) return k1.r*k1.r*rad(k2,k3)/2;
return cross(A[0],A[1])/2+k1.r*k1.r*(rad(k2,A[0])+rad(A[1],k3))/2;
}
}
db getarea(circle o,point k1,point k2,point k3){
return abs(getarea(o,k1,k2)+getarea(o,k2,k3)+getarea(o,k3,k1));
}
db r,dx,dy,x,y,p,S;
circle o;
vector<db> all;
void slove(point a,point b,point c,point d){//小矩形
db s = getarea(o,a,b,c)+getarea(o,b,c,d);
if(sign(s)>0){
all.push_back(s);
// a.print();b.print();c.print();d.print();
// printf("%.11f\n\n\n\n",s);
}
}
int main(){
scanf("%lf%lf%lf%lf%lf%lf",&r,&dx,&dy,&x,&y,&p);
o.o={0, 0};o.r=r;S = dx*dy;
x+=dx*10000;
y+=dy*10000;
while (x>-r)x-=dx;while (y>-r)y-=dy;
db minx = x,miny = y;
point n;
for(db i=minx;i<r;i+=dx){
for(db j=miny;j<r;j+=dy){
n={i,j};
slove(n,n+(point){dx,0},n+(point){0,dy},n+(point){dx,dy});
}
}
db mx = *max_element(all.begin(),all.end());
int ans=0;
// printf("%d\n",(int)all.size());
// sort(all.begin(),all.end());
for(auto x:all){
// printf("%.11f\n",x);
if(mx*p+eps>=x)ans++;
}
printf("%d\n",ans);
}