Problem 2016 How Long Is it
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
You are given two points in earth. (Points are described as the latitude (between -90 and +90 degrees) and longitude (between -180 and +180 degrees)) You are expected to calculate the distance between them ( in km). In this problem the radius of earth is 6378 km.
Input
In the first line one integer T indicates the number of test cases. (T <= 1000)
For each case, only two points each giving the latitude (between -90 and +90 degrees) and longitude (between -180 and +180 degrees) .
Output
For each test case, output the distance between them, the answer should accurate to 0.01.
Sample Input
4
-90 180 -90 90
-90 180 -90 90
-90 0 -90 90
-90 0 90 90
Sample Output
0.00
0.00
0.00
20037.08
Source
FOJ有獎月賽-2011年03月
水題。。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4
5 using namespace std;
6
7 const double PI = 3.141592653589793;
8 const double R = 6378;
9
10 #define A(x) ( (x) * PI / 180.0 )
11
12 int main() {
13 int td, a, b, c, d;
14 double da, db, dc, dd, px, py, pz, qx, qy, qz, ang, dist;
15 scanf( "%d", &td );
16 while ( td-- > 0 ) {
17 scanf( "%d%d%d%d", &a, &b, &c, &d );
18 da = A(a);
19 db = A(b);
20 dc = A(c);
21 dd = A(d);
22
23 pz = R * sin( da );
24 qz = R * sin( dc );
25
26 px = R * cos( da ) * cos( db );
27 qx = R * cos( dc ) * cos( dd );
28
29 py = R * cos( da ) * sin( db );
30 qy = R * cos( dc ) * sin( dd );
31
32 ang = acos( (px*qx+py*qy+pz*qz) /
33 sqrt(px*px+py*py+pz*pz) /
34 sqrt(qx*qx+qy*qy+qz*qz) );
35 dist = R * ang;
36
37 printf( "%0.2lf\n", dist );
38 }
39 return 0;
40 }
41