[UVA] 10693 - Traffic Volume
Problem F
Traffic
Volume
Input: standard input
Output: standard output
Time Limit: 1 second
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 v²/(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.
Input
The 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/second². The input is terminated by a single line whose value of L and f is zero.
Output
For 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
5 3 0 0 |
5.47722558 1971.80120702
|
Problem setter: Shahriar Manzoor, Member of Elite Problemsetters' Panel
Special Thanks: Derek Kisman
計算公式 !
f(v) = 3600 * v / (L + d) //流量公式, 記得換算單位
d = v*v / (2f)
f(v) = 3600 * v / (L + v*v/(2f))
f'(v) = (4fL - 2fv*v) / (2Lf + v*v)^2
得 v = sqrt(2fL)
#include <stdio.h>
#include <math.h>
int main() {
double L, f;
while(scanf("%lf %lf", &L, &f) == 2) {
if(L == 0 && f == 0)
break;
double v = sqrt(2*L*f);
double fun = 3600*v/(L+v*v/2/f);
printf("%.8lf %.8lf\n", v, fun);
}
return 0;
}
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