POJ 2208 已知边四面体六个长度，计算体积

全栈程序员-站长 • 2022年1月6日上午11:00 • 未分类 • 阅读 42

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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 5

Sample 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;
}

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