B. Ball
In Geometry, a spherical cap is a portion of a sphere cut off by a plane. If the radius of the sphere is \(r\) and the height of the spherical cap is \(h\), the volume of the spherical cap is \(πh^{2}(r−\frac{1}{3}h)\).

There are two spheres in the three-dimensional space whose radius are \(r_1\)

Input
The first line contains four integers, denoting \(x1, y1, z1, r1(0 \leq x1,y1,z1,r1\leq 100)\).

The second line contains four integers, denoting \(x2, y2, z2, r2(0 \leq x2,y2,z2,r2 \leq 100)\)

Output
Output one line with one real number, the volume of their combination.

Your answer will be consider correct if its absolute or relative error will not exceed \(10^{−6}\).

Examples

inputCopy
0 0 0 4
3 4 0 3
outputCopy
361.911473693544167

inputCopy
0 0 0 5
3 0 0 3
outputCopy
538.521340702850352

求兩個球體並起來的體積。

首先分三種情況：包含，分離，相交。

前兩種利用球體公式，後一種用題目給的球冠公式做。

#include<bits/stdc++.h>
using namespace std;

const long double pi = acos(-1);

long double getans(long double r,long double h){
    return pi * h * h * (3 * r - h);
}
long double get(long double r){
    return 4 * pi * r * r * r;
}

long double ax,ay,az,bx,by,bz,ar,br;

int main(){
    cin >> ax >> ay >> az >> ar;
    cin >> bx >> by >> bz >> br;
    
    long double dis = (ax - bx) * (ax - bx) + (ay - by) * (ay - by) + (az - bz) * (az - bz);
    
    if(dis >= (ar + br) * (ar + br)){
        printf("%.15Lf\n",(get(ar) + get(br)) / 3.0);
        return 0;
    }
    
    dis = sqrt(dis);
    
    if(dis + ar <= br || dis + br <= ar){
        printf("%.15Lf\n",get(max(ar,br))/3.0);
        return 0;
    }
    
    long double x = ar * ar - br * br + dis * dis; x = x / (2 * dis); 
    long double y = dis - x;
    
    printf("%.15Lf\n",(getans(ar,ar + x) + getans(br,br + y)) / 3.0);
    return 0;
}