1433 - Minimum Arc Distance PDF (English) Statistics ForumTime Limit: 2 second(s) Memory Limit: 32 MBYou all probably know how to calculate the distance between two points in two dimensional cartesian plane. But in this problem you have to find the minimum arc distance between two points and they are on a circle centered at another point.
You will be given the co-ordinates of the points A and B and co-ordinate of the center O. You just have to calculate the minimum arc distance between A and B. In the picture, you have to calculate the length of arc ACB. You can assume that A and B will always be on the circle centered at O.
InputInput starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with a line containing six integers Ox, Oy, Ax, Ay, Bx, By where (Ox, Oy) indicates the co-ordinate of O, (Ax, Ay) denote the co-ordinate of A and (Bx, By) denote the co-ordinate of B. All the integers will lie in the range [1, 10000].
OutputFor each case, print the case number and the minimum arc distance. Errors less than 10-3 will be ignored.
Sample InputOutput for Sample Input55711 3044 477 2186 3257 77463233 31 3336 1489 1775 134453 4480 1137 6678 2395 57168757 2995 4807 8660 2294 54294439 4272 1366 8741 6820 9145Case 1: 6641.81699183Case 2: 2295.92880Case 3: 1616.690325Case 4: 4155.64159340Case 5: 5732.01250253
#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; const double pi = 3.1415926535898; double dist(double x1,double y1,double x2,double y2){ return sqrt(pow(x1-x2,2)+pow(y1-y2,2)); } int main(){ int T; double ox,oy,ax,ay,bx,by,a,b,c,arc; scanf("%d",&T); for(int t=1;t<=T;t++){ scanf("%lf%lf%lf%lf%lf%lf",&ox,&oy,&ax,&ay,&bx,&by); a = dist(ox,oy,ax,ay); c = dist(ax,ay,bx,by); arc = acos((a*a*2-c*c)/(2*a*a)); printf("Case %d: %.3lf\n",t,a*arc); } return 0; }
转载于:https://www.cnblogs.com/yuanshixingdan/p/5563800.html
