Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2237 Accepted Submission(s):
859
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?
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.
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
分析：这个题就是看车子是否能通过这个转角！
如图分析，如果h最大的时候还是小于y的，那么车一定能通过！
即变相的求函数h=l*sin(a)-x*tan(a)+d/cos(a)的最大值，显然用三分法！
1 #include<stdio.h> 2 #include<math.h> 3 #define PI 3.1415926535 4 double l,d,x,y; 5 double hs(double p) 6 { 7 double w; 8 w=l*sin(p)-x*tan(p)+d/cos(p);//函数，计算h 9 return w; 10 } 11 int main() 12 { 13 while(scanf("%lf%lf%lf%lf",&x,&y,&l,&d)!=EOF) 14 { 15 double a=0,b=PI/2,mid,midmid; 16 do 17 { 18 //mid=(b-a)/3+a; 19 //midmid=(b-a)*2/3+a; 20 mid=(a+b)/2; 21 midmid=(mid+b)/2; 22 if(hs(mid)>hs(midmid)) 23 b=midmid; 24 else 25 a=mid; 26 }while(b-a>1e-10); 27 if(hs(mid)<y) 28 printf("yes "); 29 else 30 printf("no "); 31 } 32 return 0; 33 }