Problem Description

Mr. West bought a new car! So he is travelling around the city.

One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d.

Can Mr. West go across the corner?

 

Input

Every line has four real numbers, x, y, l and w.

Proceed to the end of file.

 

Output

If he can go across the corner, print "yes". Print "no" otherwise.

 

Sample Input

10 6 13.5 410 6 14.5 4

 

Sample Output

yesno

 

 Idea

To begin with, if d is smaller than x or y, then it's sure that the car couldn't pass the conner. Besides, we can considerate the l, which can be composed by a.

l = [(x / sin(a) + y / cos(a)) * cos(a) - d / sin(a)] / cos(a)

simplify the equality, we can get

l = x / sin(a) + y / cos(a) - d / sin (a) / cos(a)

then I can get l' and l'', but I can't find the any regulation about those.

what's a pity!

so I just use trichotomy to get similarity of l. I know it's luck.

 

Code

1 #include 
         
           2 #include 
          
            3 double x, y, l, d; 4 double f(double a) 5 { 6     return x / sin(a) + y / cos(a) - d / sin(a) / cos(a); 7 } 8 void main() 9 {10     double low, high, mid1, mid2;11     int i;12     while (scanf("%lf %lf %lf %lf", &x, &y, &l, &d) != EOF)13     {14         //there are two situation15         if (d >= x || d >= y)16         {17             puts("no");18         }19         else20         {21             high = acos(-1.0) / 2;22             low = 0;23             while (fabs(high - low) >= 1e-7)24             {25                 mid1 = (high + low) / 2;26                 mid2 = (mid1 + high) / 2;27                 //printf("%f, %f, %f, %f\n",28                 //        mid1, mid2, f(mid1), f(mid2));29                 if (f(mid1) < f(mid2))30                     high = mid2;31                 else32                     low = mid1;33             }34             //printf("high = %f, low = %f\n", high, low);35             if (l < f(low))36                 puts("yes");37             else38                 puts("no");39         }40     }41 }
          
         

 

 

Key Points

Pay more attention to the algorithm.