UVa 10693 Traffic Volume （数学&物理模型）2014-04-26 csdn博客 synapse710693 - Traffic VolumeTime limit: 3.000 secondshttp://uva.onlinejudge.org/index.php？ option=com_onlinejudge&Itemid=8&category=467&page=show_problem&problem=16 34
In the picture below （or above depending on HTML response :）） you can see a street. It has infinite number of cars on it. The distance between any two consecutive cars is
d, length of each car is
L and the velocity of each car is
v. The volume of cars through a road means the number of cars passing through a road in a specific amount of time. When the velocity is constant,
d must be minimum for the volume of cars passing through the road to be maximal. In our model when the velocity of all the cars is
v then the minimum possible value of
d is
v/（2f） （The more the car velocity the more distance you need to bring down your velocity to zero）. Here
f is the deceleration due to break.Keeping this model in mind and given the value of
L and
f your job is to find the value of
v for which the volume of traffic through the road is maximal.InputThe input file contains several lines of input. Each line of input contains two integers
L （0<L<=100） and
f（0<f<=10000）. The unit of
L is meter and the unit of
f is meter/second. The input is terminated by a single line whose value of
L and
f is zero.OutputFor each line of input except the last one produce one line of output. Each line contains two floating-point number
v and
volume separated by a single space. Here
v is the velocity for which traffic flow is maximal and
volume is the maximum number of vehicles （of course it is a fraction） passing through the road in an hour. These two floating points should have eight digits after the decimal. Errors less than
1e-5 will be ignored.
Sample Input Output for Sample Input
首先要使每辆车通过某个点的所用时间最少，即（l+v*v/2f）/v=l/v+v/2f最少，所以当v=sqrt（2fl） 时满足题意。完整代码：
/*0.009s*/#include<cstdio>#include<cmath>int main（void）{int l, f;while （scanf（"%d%d", &l, &f）, l）{f <<= 1;printf（"%f %f
", sqrt（l * f）, 1800 * sqrt（（double）f / l））;}return 0;} 
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