POJ 2208--Pyramids(欧拉四面体体积计算)

Pyramids

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 3448 Accepted: 1122 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

1999.9938

Source

Northeastern Europe 2002, Northern Subregion

利用欧拉四面体计算公式带入计算

和HDU 1411一样,只是是否是多组输入的问题。

/*
*Memory: 188K       Time: 47MS
*Language: C++      Result: Accepted
*/
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
//p,q,r为AD,BD,CD;n,l,m为AB,BC,AC;
double V_tetrahedron(double l, double m, double n, double q, double p, double r) {
    //return sqrt((4 * p*p*q*q*r*r - p*p*(q*q + r*r - l*l)*(q*q + r*r - l*l) - q*q*(r*r + p*p - m*m)*(r*r + p*p - m*m) - r*r*(p*p + q*q - n*n)*(p*p + q*q - n*n) + (p*p + q*q - n*n)*(q*q + r*r - l*l)*(r*r + p*p - m*m))) / 12.0;
    double x = q*q + r*r - l*l, y = r*r + p*p - m*m, z = p*p + q*q - n*n;
    return sqrt((4 * p*p*q*q*r*r - p*p*x*x - q*q*y*y - r*r*z*z + z*x*y)) / 12.0;
}
int main(){
    double l,m,n,q,p,r;
    cin>>n>>m>>p>>l>>q>>r;
    double ans=V_tetrahedron(l,m,n,q,p,r);
    printf("%.4lf\n",ans);
    return 0;
}