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 descriptionlong 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

Inputmulti 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]

<R, HR, Hz <= 100

Outputfor each case, output the probability of the war which happens between A and B. The answer shoshould accurate to six decimal places.

Sample Input

1 1 12 1 1

Sample output

0.6666670.187500

Sourcebestcoder round #12

Recommendheyang | we have carefully selected several similar problems for you: 5061 5059 5058 5053 question: give you a ball at the origin. A ball with a radius of R and a cylinder. The radius of the cylinder is HR, the height is Hz, and the diameter is Z axis. Then, the midpoint of the diameter is also at the origin. Then ask the volume VC/volume of the intersection and Vu. Idea: This question is easier because the point at which the origin is located. I have not written any questions about computational ry before, because the last question won't. Only the scalp is on. We will discuss the classification. R, and HR size. Then, we will discuss whether the cylinder has passed through the sphere. That is, the size of SQRT (R * r-HR * hr) and R. Use a fixed integral for the volume of a part of the sphere. The product is pi * r * z-Pi * z/3 | upper and lower limits. For details, see the code:

#include<algorithm>#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>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<r) { d=sqrt(r*r-hr*hr); if(hz<=d) vc=2*PI*hr*hr*hz; else { hh=min(hz,r); a=PI*r*r*hh-PI*hh*hh*hh/3; b=PI*r*r*d-PI*d*d*d/3; vc=2*(PI*hr*hr*d+a-b); } } else { if(hz<=r) vc=2*(PI*r*r*hz-PI*hz*hz*hz/3); else vc=4*PI*r*r*r/3; } vu=4*PI*r*r*r/3+PI*hr*hr*hz*2-vc; //printf("%lf %lf\n",vc,vu); printf("%.6lf\n",vc/vu); } return 0;}

Bestcoder round #12 war (Computational ry)