【codeforces 29B】Traffic Lights

B. Traffic Lights
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

A car moves from point A to point B at speed v meters per second. The action takes place on the X-axis. At the distance d meters from A there are traffic lights. Starting from time 0, for the first g seconds the green light is on, then for the following r seconds the red light is on, then again the green light is on for the g seconds, and so on.

The car can be instantly accelerated from 0 to v and vice versa, can instantly slow down from the v to 0. Consider that it passes the traffic lights at the green light instantly. If the car approaches the traffic lights at the moment when the red light has just turned on, it doesn't have time to pass it. But if it approaches the traffic lights at the moment when the green light has just turned on, it can move. The car leaves point A at the time 0.

What is the minimum time for the car to get from point A to point B without breaking the traffic rules?
Input

The first line contains integers l, d, v, g, r (1 ≤ l, d, v, g, r ≤ 1000, d < l) — the distance between A and B (in meters), the distance from A to the traffic lights, car's speed, the duration of green light and the duration of red light.
Output

Output a single number — the minimum time that the car needs to get from point A to point B. Your output must have relative or absolute error less than 10 - 6.

Sample test(s)

Input

2 1 3 4 5

Output

0.66666667

Input

5 4 3 1 1

Output

2.33333333

題意：車要經過一段長爲L 路，距離起點D的的地方有交通燈，起始爲綠燈，經過g秒以後變爲紅燈，紅燈亮r秒以後 變成綠燈然後循環。。。車子可以瞬間達到速度v，也可以瞬間由v變成0.問，車子經過這段路所花的最短時間。

 解題思路：數學題，在紙上畫畫，分兩種情況，一個是，恰好是綠燈，另一種就是紅燈的情況。我在一組臨界數據wa了，就是當恰好遇到綠燈將要變成紅的的時間點不能通過，需要變成另外一種情況。

code：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
using namespace std;
int main()
{
    int  l,d,v;
    int g,r;
    double ans;
    int cnt=0;
    scanf("%d%d%d%d%d",&l,&d,&v,&g,&r);
    double pre=d*1.0/v;
    while(pre>(g+r)){
        pre-=(g+r);
        cnt++;
    }
    if(pre<g)
        printf("%.8lf\n",l*1.0/v);
    else if(pre<=g+r){
        ans=(l-d)*1.0/v+(cnt+1)*(g+r);
        printf("%.8lf\n",ans);
    }

    return 0;
}