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.
Proceed to the end of file. |
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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 |
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Sample Output
yes 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：
#include<stdio.h>
#include<string.h>
#include<math.h>
const double pi=*asin(1.0);
double x,y,l,d;
double geth(double an){
double s,h;
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
h=s*tan(an)+d*cos(an);
return h;
}
double ABS(double a){
return a>=?a:-a;
}
void erfen(){
double l=,m,mm,r=pi/;
// printf("%lf\n",pi);
while(ABS(r-l)>1e-){
m=(l+r)/;
mm=(m+r)/;
if(geth(m)>=geth(mm))r=mm;
else l=m;
}
// printf("%lf\n",geth(m));
if(geth(l)>y)puts("no");
else puts("yes");
}
int main(){
while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&d)){
erfen();
}
return ;
}