sgu 151 Construct a triangle

来源：互联网发布：中国上古史知乎编辑：程序博客网时间：2024/06/05 12:04

题目描述：

151. Construct a triangle

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

input: standard input
output: standard output

Find coordinates of any triangle ABC if it is know that |AB|=c, |AC|=b, |AM|=m, AM is a median of triangle.

Input

There are three real numbers in input: c, b, m (0<c,b,m<=10^3) separated by a space. Length of the fractional part of each number is not greater than 2 digits.

Output

If solution exists, write three lines. Write coordinates of point A to first line, coordinates of B to second line and coordinates of C to third line. Separate numbers by a space; absolute value of each coordinate must not exceed 10^4. Write numbers with 5 digits after decimal point. If there is no solution, write "Mission impossible"

Sample test(s)

Input

5 5 3

Output

0.00000 3.00000
-4.00000 0.00000
4.00000 0.00000

把一个点放在原点，一个点放在(c,0) 点

那么中线长度就是

sqrt( c^2 + b^2 + 2 *cos( sita ) * b * c ) = m;

可以求出 sita ；

一切就ok了。

注意此题的三角形比较奇葩，可以是面积为0 的，也就是可以共线。

贴代码：

#include<iostream>#include<cstring>#include<cstdio>#include<set>#include<algorithm>#include<vector>#include<cstdlib>#include<cmath>#define inf 0xfffffff#define CLR(a,b) memset((a),(b),sizeof((a)))#define FOR(a,b) for(int a=1;a<=(b);(a)++)using namespace std;int const nMax = 1010;int const base = 10;typedef int LL;typedef pair<LL,LL> pij;//    std::ios::sync_with_stdio(false);double const eps = 1e-9;double const pi = acos(-1.0);double a,b,c,aa,ab;double he,ta;double F(double d){    return (aa+ab*cos(d))/4.0;}int main(){    scanf("%lf%lf%lf",&a,&b,&c);    aa=a*a+b*b;    ab=a*b*2.0;    c*=c;    double s=(4.0*c-aa)/ab;    if(fabs(s)>1){        puts("Mission impossible");        return 0;    }    he=acos(s);    printf("%.5lf %.5lf\n",0.0,0.0);    printf("%.5lf %.5lf\n",a,0.0);    printf("%.5lf %.5lf\n",b*cos(he),b*sin(he));    return 0;}