Codeforces AIM Tech Round (Div. 2) 624A A. Save Luke

A. Save Luke

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Luke Skywalker got locked up in a rubbish shredder between two presses. R2D2 is already working on his rescue, but Luke needs to stay alive as long as possible. For simplicity we will assume that everything happens on a straight line, the presses are initially at coordinates 0 and L, and they move towards each other with speed v1 and v2, respectively. Luke has width d and is able to choose any position between the presses. Luke dies as soon as the distance between the presses is less than his width. Your task is to determine for how long Luke can stay alive.

Input

The first line of the input contains four integers d, L, v1, v2 (1 ≤ d, L, v1, v2 ≤ 10 000, d < L) — Luke’s width, the initial position of the second press and the speed of the first and second presses, respectively.

Output

Print a single real value — the maximum period of time Luke can stay alive for. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let’s assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

Sample test(s)

input

2 6 2 2

output

1.00000000000000000000

input

1 9 1 2

output

2.66666666666666650000

Note

In the first sample Luke should stay exactly in the middle of the segment, that is at coordinates [2;4], as the presses move with the same speed.

In the second sample he needs to occupy the position . In this case both presses move to his edges at the same time.

题意：

给出四个输入:d,L,v1,v2。大意是在横坐标最长为L的横轴上，敌人R1以速度v1从坐标0出发，R2以速度v2从坐标L出发，Luke的交通工具长度为d，问敌人同时向中间夹击的情况下，Luke最长能活多久。

思路：

有点像高中物理题有没有！把敌人的速度抽象为同方向，Luke要活最久，那么必须离敌人最远，故敌人必须在坐标L的位置，Luke要在坐标0的位置，则敌人的总速度v = v1 + v2，并从L向坐标d出发，这样抽象理解的话就很好做了。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
double d,l,v1,v2;
while(cin>>d>>l>>v1>>v2)
{
double cnt = 0.0;
double v = v1 + v2;
for(double i = 1;;i++)
{
double dis = v * i;
double flag = l - dis - d;
if(flag <= v)
{
cnt = i;
cnt += flag / v;
break;
}
}
printf("%.6f\n",cnt);
}
return 0;
}