HDOJ_ACM_Turn the corner

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 4
10 6 14.5 4

 

Sample Output

yes
no

 

 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

#include <stdio.h>
#include <math.h>
double x, y, l, d;
double f(double a)
{
    return x / sin(a) + y / cos(a) - d / sin(a) / cos(a);
}
void main()
{
    double low, high, mid1, mid2;
    int i;
    while (scanf("%lf %lf %lf %lf", &x, &y, &l, &d) != EOF)
    {
        //there are two situation
        if (d >= x || d >= y)
        {
            puts("no");
        }
        else
        {
            high = acos(-1.0) / 2;
            low = 0;
            while (fabs(high - low) >= 1e-7)
            {
                mid1 = (high + low) / 2;
                mid2 = (mid1 + high) / 2;
                //printf("%f, %f, %f, %f\n",
                //        mid1, mid2, f(mid1), f(mid2));
                if (f(mid1) < f(mid2))
                    high = mid2;
                else
                    low = mid1;
            }
            //printf("high = %f, low = %f\n", high, low);
            if (l < f(low))
                puts("yes");
            else
                puts("no");
        }
    }
}

 

Key Points

Pay more attention to the algorithm.