POJ 2208 已知边四面体六个长度，计算体积
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Pyramids
Time Limit: 1000MS | Memory Limit: 65536K | |||
|---|---|---|---|---|
Total Submissions: 2718 | Accepted: 886 | Special Judge | ||
Description
Recently in Farland, a country in Asia, a famous scientist Mr. Log Archeo has discovered ancient pyramids. But unlike those in Egypt and Central America, they have triangular (not rectangular) foundation. That is, they are tetrahedrons in mathematical sense. In order to find out some important facts about the early society of the country (it is widely believed that the pyramid sizes are in tight connection with Farland ancient calendar), Mr. Archeo needs to know the volume of the pyramids. Unluckily, he has reliable data about their edge lengths only. Please, help him!
Input
The file contains six positive integer numbers not exceeding 1000 separated by spaces, each number is one of the edge lengths of the pyramid ABCD. The order of the edges is the following: AB, AC, AD, BC, BD, CD.
Output
A real number — the volume printed accurate to four digits after decimal point.
Sample Input
1000 1000 1000 3 4 5Sample Output
依据边来求出四面体的高，然后公式计算。
代码：
/* ***********************************************
Author :_rabbit
Created Time :2014/5/9 21:32:01
File Name :5.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
double volume(double a,double b,double c,double d,double e,double f){
double a2=a*a,b2=b*b,c2=c*c,d2=d*d,e2=e*e,f2=f*f;
double tr1=acos((c2+b2-f2)/(2*b*c));
double tr2=acos((a2+c2-e2)/(2*a*c));
double tr3=acos((a2+b2-d2)/(2*a*b));
double tr4=(tr1+tr2+tr3)/2;
double temp=sqrt(sin(tr4)*sin(tr4-tr1)*sin(tr4-tr2)*sin(tr4-tr3));
return a*b*c*temp/3;
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
double a,b,c,d,e,f;
while(cin>>a>>b>>c>>d>>e>>f)printf("%.4f\n",volume(a,b,c,d,e,f));
return 0;
}版权声明：本文博客原创文章。博客，未经同意，不得转载。
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