Using this information along with how much gasoline is available for your planet-rover (which gets a measley 5 miles per gallon), you have to determine if you can possibly get to the crash site and back without running out of fuel.
注意题目所说的是来回车程,而且所给的角不一定就是车所走的路线对应的角。。还需判断若角度大于180度,则应该选择小于180的角走。。
#include<iostream>
#include<string.h>
using namespace std;
#define PI 3.14159
int main()
{
int r,y,o,v=5;
char s[10];
float l;
while(cin>>s)
{
if(strcmp(s,"START")==NULL)
{
cin>>r>>y>>o>>s;
if(o>180)o=360-o;
l=o*1.0/360*2*PI*r;
if(v*y>=2*l)cout<<"YES "<<(int)(v*y-2*l)/v<<endl;
else
cout<<"NO "<<v*y<<endl;
}
else
if(strcmp(s,"ENDOFINPUT")==NULL)break;
}
return 0;
}