hdu6158

201708 （ACM）计算几何----圆反演

• 反演做法

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define mp make_pair
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define de(a) cout<<#a<<"="<<a<<endl;
typedef long long ll;
typedef double db;
typedef pair<int, int> pii;
typedef vector<int> vi;
//--------
const db eps=1e-13;
const db pi=acos(-1);
int r1,r2,n;
struct P {
    db x,y;
    P() {}
    P(db x,db y) {
        this -> x = x;
        this -> y = y;
    }
    P operator - (const P &c) const {
        return P(x - c.x, y - c.y);
    }
};
db dis(P a) {
    return sqrt(a.x*a.x+a.y*a.y);
}
db dis(P a,P b) {
    return dis(a-b);
}
db gao(db x) {
    return 1/x;
}
int main() {
    int T;scanf("%d",&T);
    while(T--) {
        scanf("%d%d%d",&r1,&r2,&n);
        if(r1<r2) swap(r1,r2);
        db x1=gao(2*r1),x2=gao(2*r2);
        db d=x2-x1;
        db ans=0,r=0;
        P o = P(0, 0);
        rep(i,1,n+1) {
            if(i==1) {
                r=r1-r2;
            } else if(i%2==0) {
                P c = P(x1+d/2, d*i/2);
                db dc = dis(c, o);
                db da=dc-d/2,d_=dc+d/2;
                db a=gao(da),b=gao(d_);
                r=fabs(a-b)/2;
            }
            ans+=r*r;
            if(r*r<eps) break;
        }
        printf("%.5f\n",ans*pi);
    }
    return 0;
}