Turn the corner |
| Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 151 Accepted Submission(s): 61 |
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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? |
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Input Every line has four real numbers, x, y, l and w. |
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Output If he can go across the corner, print "yes". Print "no" otherwise. |
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Sample Input 10 6 13.5 4 10 6 14.5 4 |
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Sample Output yes no
汽车拐弯问题，给定X, Y, l, d判断是否能够拐弯。 我们发现随着角度θ的增大，最大高度ｈ先增长后减小，即为凸性函数，可以用三分法来求解。 这里的Calc函数需要比较繁琐的推倒公式： 3分搜索法 |
代码：
1 #include<stdio.h>
2 #include<string.h>
3 #include<math.h>
4 const double pi=2*asin(1.0);
5 double x,y,l,d;
6 double geth(double an){
7 double s,h;
8 s=l*cos(an)-x+d*sin(an);//应该是减x因为车头要对住墙，然后看车尾最高是否大于y
9 h=s*tan(an)+d*cos(an);
10 return h;
11 }
12 double ABS(double a){
13 return a>=0?a:-a;
14 }
15 void erfen(){
16 double l=0,m,mm,r=pi/2;
17 // printf("%lf\n",pi);
18 while(ABS(r-l)>1e-10){
19 m=(l+r)/2;
20 mm=(m+r)/2;
21 if(geth(m)>=geth(mm))r=mm;
22 else l=m;
23 }
24 // printf("%lf\n",geth(m));
25 if(geth(l)>y)puts("no");
26 else puts("yes");
27 }
28 int main(){
29 while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&d)){
30 erfen();
31 }
32 return 0;
33 }
时间： 2024-12-20 13:07:15