zoj 2500 求两球体积并

最新推荐文章于 2021-03-16 20:34:16 发布

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最新推荐文章于 2021-03-16 20:34:16 发布

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分类专栏： acm_数学问题文章标签： output action each input

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本文介绍了一个计算两个原子间电子可能出现的空间体积的算法。输入包括两个原子的坐标和半径，输出则是精确到小数点后两位的可能空间体积。算法考虑了不同情况下原子间的相对位置及重叠状态。

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In order to describe a nuclear action, you are asked to tell where the electrons may occur.

Here we assume that the electrons only occur within a sphere whose location and radius are given. The action consists of only two atoms.

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. (0 < T ≤ 50)

Each test case consists of a line containing 8 integers X1, Y1, Z1, R1, X2, Y2, Z2 and R2, describing the coordinates and the radiuses of the two atoms. The absolute value of each coordinate do not exceed 100 and 0 < R1, R2 ≤ 100.

Output

For each test case, output the volume of space where the electrons may occur, accurated to the nearest 0.01.

Sample Input

3
0 0 0 1 0 0 2 1
0 0 0 2 0 0 2 2
0 0 0 2 0 0 2 1

Sample Output

8.38
56.55
36.00

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const double pi=acos(-1.0);
int x[2],y[2],z[2];
int R,r;
double d;
int main()
{
    int ci;scanf("%d",&ci);
    while(ci--)
    {
        scanf("%d%d%d%d%d%d%d%d",&x[0],&y[0],&z[0],&R,&x[1],&y[1],&z[1],&r);
        double d=sqrt((x[0]-x[1])*(x[0]-x[1])+(y[0]-y[1])*(y[0]-y[1])+(z[0]-z[1])*(z[0]-z[1]));
        double v1=4*pi*R*R*R/3,v2=4*pi*r*r*r/3;
        if(d>=R+r)
        {
            printf("%.2lf/n",v1+v2);
        }
        else if(fabs(R-r)>=d)
        {
            printf("%.2lf/n",max(v1,v2));
        }
        else
        {
            if(R<r) swap(R,r);
            double insrt=(pi*(R+r-d)*(R+r-d)*(d*d+2*d*r-3*r*r+2*d*R+6*r*R-3*R*R))/(12*d);
            printf("%.2lf/n",v1+v2-insrt);
        }
    }
    return 0;
}