codeforces 5D

D. Follow Traffic Rules

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Everybody knows that the capital of Berland is connected to Bercouver (the Olympic capital) by a direct road. To improve the road's traffic capacity, there was placed just one traffic sign, limiting the maximum speed. Traffic signs in Berland are a bit peculiar, because they limit the speed only at that point on the road where they are placed. Right after passing the sign it is allowed to drive at any speed.

It is known that the car of an average Berland citizen has the acceleration (deceleration) speed of a km/h², and has maximum speed of v km/h. The road has the length of l km, and the speed sign, limiting the speed to w km/h, is placed d km (1 ≤ d < l) away from the capital of Berland. The car has a zero speed at the beginning of the journey. Find the minimum time that an average Berland citizen will need to get from the capital to Bercouver, if he drives at the optimal speed.

The car can enter Bercouver at any speed.

Input

The first line of the input file contains two integer numbers a and v (1 ≤ a, v ≤ 10000). The second line contains three integer numbers l, d and w (2 ≤ l ≤ 10000; 1 ≤ d < l; 1 ≤ w ≤ 10000).

Output

Print the answer with at least five digits after the decimal point.

Examples

Input

1 1
2 1 3

Output

2.500000000000

Input

5 70
200 170 40

Output

8.965874696353

吐槽：这真是一道有意思的题。。纯高中物理题用代码实现的感觉就是不一样呀

题意：一辆车初速度为0，以恒定的加速度a在长为l的公路上行驶，据起点d处有一测速点，要求车到达此点时速度不得超过w，求走完这条公路的最短时间。

解题思路：跟高中做物理题一样，一点一点分析。
下面是我的分析：
一 v<=w
　　1.一直加速到终点
　　2.加速到v后匀速到终点
二 v>w
　　1.在d段中一直加速也到不了w，那就一直加速
　　2.在d段中加速超过w到达v1，然后减速在d点刚好速度为w
　　3.在d段中加速超过w到达v，接着匀速一段，然后减速在d点刚好速度为w

　 以上三种情况到达d后，再按照一中两个情况分析（l-d)的路程。

附ac代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <stack>
#include <cmath>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    double a,v,l,d,w;
    scanf("%lf%lf%lf %lf%lf",&a,&v,&l,&d,&w);
    double ans=0,t1=0,t2=0,x1=0,x2=0;
    double det=0;
    if(v<=w) {
        t1=v/a;
        x1=0.5*a*t1*t1;
        if(x1>=l) {
            ans=sqrt(2*l/a);
        }
        else {
            x2=l-x1;
            t2=x2/v;
            ans=t1+t2;
        }
    }
    else {
        t1=sqrt(2*d/a);
        if(t1*a<=w) {
            t1=v/a;
            x1=0.5*a*t1*t1;
            if(x1>=l) {
                ans=sqrt(2*l/a);
            }
            else {
                x2=l-x1;
                t2=x2/v;
                ans=t1+t2;
            }
        }
        else {
            det=sqrt(2*w*w+4*a*d);
            t2=(det-2*w)/(2*a);
            if(w+a*t2<=v){
                t1=(w+a*t2)/a;
                ans=t1+t2;
            }
            else {
                t1=v/a;
                t2=(v-w)/a;
                x1=0.5*a*t1*t1+v*t2-0.5*a*t2*t2;
                x2=d-x1;
                ans=t1+t2+x2/v;
            }
            
            t1=(v-w)/a;
            x1=w*t1+0.5*a*t1*t1;
            if(x1>(l-d)) {
                det=sqrt(w*w+2*a*(l-d));
                ans+=(det-w)/a;
            }
            else {
                ans+=t1;
                x2=l-d-x1;
                ans+=(x2/v);
            }
        }
    }
        printf("%.12lf
",ans);
    return  0;
}