1294. Mars Satellites
Time limit: 1.0 second
Memory limit: 64 MB
Four artificial satellites travel in one plane along the areostationary orbit around Mars. They have code names A, B, C and D and travel exactly in this order. Venus’s scouts for military purposes (for what particular purpose they did not say) decided to find
a distance between satellites C and D. All Mars satellites could measure distances to the other satellites, that is why all what is needed to do is to penetrate in the computer system of satellite C and measure the distance to satellite D (or vice versa).
Nevertheless, Martians are not so stupid and have not very bad defense. That is why all what could Venus’s scouts do is to break the defense of satellites A and B (that were older models). They measured distances from satellites A and B to satellites C and
D, but now they do not know how to find the distance from C to D using these measurements. You can help them.
Input
There are 4 numbers: distances from A to D, from A to C, from B to D and from B to C in thousands kilometers (integers from 1 to 10000). Satellites can measure distance even through the planet and you may assume that orbit is a circle. Do not assume the radius
of the orbit equal to 20392 km as it should be for the real areostationary orbit.
Output
If it is impossible to find out the distance from C to D with these data, you should print "Impossible.", otherwise you are to print "Distance is X km.", where X is the required distance in kilometers (rounded to the integer number).
Sample
input
4 7 5 7
output
Distance is 5385 km.
题意：有ABCD四个点在圆上，按顺序排列（也就是说B一定在AC之间）。 然后根据输入的四条边。
做法：因为同弦的圆周角相等，所以∠A==∠B。 然后在三角形ADC和三角形BCD中各用一遍余弦定理。可以得到两个方程。未知的只有cos∠A 和 DC，一个二元方程组，化简一下 就可以求出DC了。
余弦定理 a^2=b^2+c^2-2*b*c*cos(∠A);
#pragma warning (disable:4786)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64d
#define pi acos(-1.0)
int main()
{
double zuo,you,c,ans,AD,AC,BD,BC;
while(scanf("%lf%lf %lf%lf",&AD,&AC,&BD,&BC)!=EOF)
{
if(AD*AC==0||BD*BC==0)
{
puts("Impossible.");
continue;
}
zuo=(AD*AD+AC*AC)/(2.0*AD*AC);
you=(BD*BD+BC*BC)/(2.0*BD*BC);
c=(2*BD*BC-2*AD*AC)/(2*BD*BC*2*AD*AC);
if(c==0||(zuo-you)/c<0)
{
puts("Impossible.");
continue;
}
ans=sqrt((zuo-you)/c);
printf("Distance is %.0lf km.\n",ans*1000);
}
return 0;
}