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POJ 3407 求球面距离
题意：给出球面上两点的经纬度，注意分跟秒进制是60进制的，让求球面上的两点距离。
有公式 r*acos(sin(a1)*sin(a2)+cos(a1)*cos(a2)*cos(b1-b2)) 。
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
double Dis3D(double a1,double b1,double a2,double b2,double r)
{
return r*acos(sin(a1)*sin(a2)+cos(a1)*cos(a2)*cos(b1-b2));
}
double geta(double x1,double x2,char *s)
{
double a=x1+x2/60.0;
if(s[0]=='W'||s[0]=='S')
a=-a;
a=a*acos(-1.0)/180;
return a;
}
int main()
{
double x1,x2,y1,y2;
char s1[3],s2[3];
while(~scanf("%lf%lf%s%lf%lf%s",&x1,&x2,s1,&y1,&y2,s2))
{
double a1=geta(x1,x2,s1),b1=geta(y1,y2,s2);
scanf("%lf%lf%s%lf%lf%s",&x1,&x2,s1,&y1,&y2,s2);
double a2=geta(x1,x2,s1),b2=geta(y1,y2,s2);
printf("%.3f\n",Dis3D(a1,b1,a2,b2,6370));
}
return 0;
}
最后更新：2017-04-03 16:48:37