题解

#include<cmath>
#include<queue>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef double dl;
const int N=510;
const dl eps=1e-8;
struct cir{
    dl x,r;
}p[N];
struct line{
    dl k,b,l,r;
}q[N];
int n;
dl alpha;
inline dl py(dl a,dl b){return sqrt(a*a-b*b);}
inline dl f(dl x){
    dl ans=0;
    for(int i=1;i<=n;i++){
        if(p[i].x-p[i].r<x&&x<p[i].x+p[i].r){
            ans=max(ans,py(p[i].r,x-p[i].x));
        }
    }
    for(int i=1;i<n;i++){
        if(q[i].l<=x&&x<=q[i].r){
            ans=max(ans,q[i].k*x+q[i].b);
        }
    }
    return ans;
}
inline dl simpson(dl l,dl r){
    dl mid=(l+r)/2;
    return (f(l)+4*f(mid)+f(r))*(r-l)/6;
}
inline dl asr(dl l,dl r,dl ans){
    dl mid=(l+r)/2;
    dl l1=simpson(l,mid),r1=simpson(mid,r);
    if(fabs(l1+r1-ans)<eps)return l1+r1;
    return asr(l,mid,l1)+asr(mid,r,r1);
}
inline int shadow(dl a){
    dl c=a*alpha;
    dl d=sqrt(a*a+c*c);
    return c*a/d;
}
inline void getl(int x,int y){
    if(fabs(p[x].r-p[y].r)<eps){
        q[x].l=p[x].x,q[x].r=p[y].x;
        q[x].k=0;q[x].b=p[x].r;
        return;
    }
    dl CA=p[y].x-p[x].x,AJ=fabs(p[x].r-p[y].r);
    dl CG=p[x].r,AB=p[y].r;
    dl CK=CG*AJ/CA,AI=AB*AJ/CA;
    if(p[x].r>p[y].r){
        q[x].l=p[x].x+CK;q[x].r=p[y].x+AI;
        dl x1=q[x].l,y1=py(CG,CK);
        dl x2=q[x].r,y2=py(AB,AI);
        q[x].k=(y1-y2)/(x1-x2);
        q[x].b=y1-q[x].k*x1;
    }else{
        q[x].l=p[x].x-CK;q[x].r=p[y].x-AI;
        dl x1=q[x].l,y1=py(CG,CK);
        dl x2=q[x].r,y2=py(AB,AI);
        q[x].k=(y1-y2)/(x1-x2);
        q[x].b=y1-q[x].k*x1;
    }
}
int main(){
    scanf("%d%lf",&n,&alpha);
    alpha=1.0/tan(alpha);
    for(int i=1;i<=n+1;i++){
        scanf("%lf",&p[i].x);
        p[i].x*=alpha;
        p[i].x+=p[i-1].x;
    }
    for(int i=1;i<=n;i++)scanf("%lf",&p[i].r);
    p[++n].r=0;
    for(int i=1;i<n;i++){
        getl(i,i+1);
    }
    dl l=p[1].x-p[1].r,r=p[n].x;
    for(int i=1;i<=n;i++){
    r=max(r,(p[i].x+p[i].r));
    l=min(l,(p[i].x-p[i].r));
    }
    printf("%.2lf\n",2*asr(l,r,simpson(l,r)));
    return 0;
}