BestCoder Round #12 War(計算幾何)

War

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 81 Accepted Submission(s): 23
Special Judge

Problem Description Long long ago there are two countrys in the universe. Each country haves its own manor in 3-dimension space. Country A's manor occupys x^2+y^2+z^2<=R^2. Country B's manor occupys x^2+y^2<=HR^2 && |z|<=HZ. There may be a war between them. The occurrence of a war have a certain probability.
We calculate the probability as follow steps.
1. VC=volume of insection manor of A and B.
2. VU=volume of union manor of A and B.
3. probability=VC/VU
Input Multi test cases(about 1000000). Each case contain one line. The first line contains three integers R,HR,HZ. Process to end of file.

[Technical Specification]
0< R,HR,HZ<=100

Output For each case，output the probability of the war which happens between A and B. The answer should accurate to six decimal places.
Sample Input

1 1 1
2 1 1

Sample Output

0.666667
0.187500

Source BestCoder Round #12
Recommend heyang | We have carefully selected several similar problems for you: 5061 5059 5058 5053 5052 題意： 給你一個球心在原點.半徑為r的球和一個圓柱體。圓柱體半徑為hr,高為hz然後通徑為z軸.然後通徑中點也在原點。 然後問你相交部分的體積vc/體積並vu。 思路： 這題由於原點中點都在原點所以比較好做。一前還沒怎麼寫過計算幾何的題，由於最後一題不會。只有硬著頭皮上了。我們就分類討論。r,和hr的大小。然後在討論下圓柱體有沒有穿出球體。即sqrt(r*r-hr*hr)和r的大小。對於一部分的球體的體積用用定積分.積出來為PI*r*r*z-PI*z*z*z/3|上下限。 詳細見代碼：

#include
#include
#include
#include
#include
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
const double PI=acos(-1.0);
const double eps=1e-8;
typedef long long ll;

int main()
{
    double r,hr,hz,vc,vu,d,a,b,hh;

    while(~scanf("%lf%lf%lf",&r,&hr,&hz))
    {
        if(hr