The Meeting Place Can Not Be Changed（二分）

The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction.

At some points on the road there are n friends, and i-th of them is standing at the point xi meters and can move with any speed no greater than vi meters per second in any of the two directions along the road: south or north.

You are to compute the minimum time needed to gather all the n friends at some point on the road. Note that the point they meet at doesn’t need to have integer coordinate.

Input
The first line contains single integer n (2 ≤ n ≤ 60 000) — the number of friends.

The second line contains n integers x1, x2, …, xn (1 ≤ xi ≤ 109) — the current coordinates of the friends, in meters.

The third line contains n integers v1, v2, …, vn (1 ≤ vi ≤ 109) — the maximum speeds of the friends, in meters per second.

Output
Print the minimum time (in seconds) needed for all the n friends to meet at some point on the road.

Your answer will be considered correct, if its absolute or relative error isn’t greater than 10 - 6. Formally, let your answer be a, while jury’s answer be b. Your answer will be considered correct if holds.

Example
Input
3
7 1 3
1 2 1
Output
2.000000000000
Input
4
5 10 3 2
2 3 2 4
Output
1.400000000000
Note
In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.

**题意：一个数轴，给N个点，每个点能正向负向移动，给出每个点的最大速度，求这N个点最短相遇时间
首先，不知道每个点的运动方向和速度，但可以求出每个点在t内的活动区间，时间由短到长，点由不相遇到相遇， 有个临界状态，可二分时间，找出临界时间就解决了；
其中判断能否在t内每个点都相遇的方法是这ｎ个区间的交集为非空，那么ｎ个区间的左端点的最大值小于右端点的最小值，利用这个判断方法就可以了。**

注意１、ｍａｘａ和ｍｉｎｂ的确定
２、ｆｏｒ（）来进行二分，一般是二分对象是离散状态。
３、二分上下界的确定。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <iomanip>#define maxn 60010#define eps 1e-7using namespace std;typedef long long ll;int x[maxn];int  v[maxn];double a[maxn];double b[maxn];int n;bool judge(double t){    double maxa=x[0]-v[0]*t;    double minb=x[0]+v[0]*t;    for(int i=0;i<n;i++)    {        a[i]=x[i]-v[i]*t;        if(a[i]>maxa)maxa=a[i];        b[i]=x[i]+v[i]*t;        if(b[i]<minb)minb=b[i];    }    return minb-maxa>=eps;}int main(){    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%d",&x[i]);    }    for(int j=0;j<n;j++)    {        scanf("%d",&v[j]);    }    double st=0;    double ed=1000000010;    double mid;    for(int i=0;i<10000;i++)    {        mid=(st+ed)/2.0;        if(judge(mid))ed=mid;        else st=mid;    }    cout<<setprecision(8)<<st<<endl;    return 0;}