pku2208pyramids

题目：pku2208

 

题意：已知四面体的六条边长，求其体积。下面二种方法均可作为模板。

 

代码一：先求角度，再求体积。由wanglang515 编写

Memory: 176K

Time: 16MS

 

//AB, AC, AD, BC, BD, CD #include<stdio.h> #include<math.h> int main() { double a,b,c,d,e,f,v; double angle1,angle2,angle3,angle; while(scanf("%lf %lf %lf %lf %lf %lf",&a,&b,&c,&d,&e,&f)!=EOF) { angle1=acos((a*a+b*b-d*d)/(2*a*b)); angle2=acos((a*a+c*c-e*e)/(2*a*c)); angle3=acos((b*b+c*c-f*f)/(2*b*c)); angle=(angle1+angle2+angle3)/2; v=1/3.0*a*b*c*sqrt(sin(angle)*sin(angle-angle1)* sin(angle-angle2)*sin(angle-angle3)); printf("%0.4f/n",v); } return 0; }

 

代码二：先求边，再求体积。由onchf编写。

Memory: 152K

Time: 0MS

#include <stdio.h> #include <math.h> int main() { double p , s , h , v ; double xx , yy , x , y , z ; double ab , ac , ad , bc , bd , cd ; while (scanf("%lf%lf%lf%lf%lf%lf", &ab, &ac, &ad, &bc, &bd, &cd) != EOF) { xx = (ac * ac - bc * bc + ab * ab) / (ab * 2) ; yy = sqrt(ac * ac - xx * xx) ; x = (ad * ad - bd * bd + ab * ab) / (ab * 2) ; y = (bd * bd - cd * cd + (x-xx) * (x-xx) - (x-ab) * (x-ab) + yy * yy) / (yy * 2) ; z = sqrt(ad * ad - x * x - y * y) ; p = (ab + ac + bc) / 2; s = sqrt(p * (p-ab) * (p-ac) * (p-bc)) ; h = z ; v = s * h / 3 ; printf("%.4lf/n" , v) ; } return 0 ; }