codeforces The Meeting Place Cannot Be Changed

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.

题意，给出n个点 和n个点的坐标 只能往上和往下走。要求。求得所有点最短的相遇时间。
浮点数的二分，一开始用整数的二分枚举了相遇地点，后来发现错了。只用用浮点枚举二分就行。

先求出一个点的所能达到的区间。然后用别的点来收缩此区间。例如一个点往下走只能走到 z，如果z比l小，那么限制条件l应该变为z。同理。

#include <bits/stdc++.h>
using namespace std;
int a[101010];
int p[101010];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<=n;i++) cin>>p[i];
        double l=0,r=1000000000.1;
        double mid;
    int flag;
    while(r-l>1e-8)
    {
        flag=0;
         mid=(l+r)/2;
        double L=a[1]-p[1]*mid;
        double R=a[1]+p[1]*mid;
        for(int i=2;i<=n;i++)
        {
            L=max(L,a[i]-p[i]*mid);
            R=min(R,a[i]+p[i]*mid);
            if(L>R)
            {
                flag=1;
                break;
            }
        }
        if(flag)
        {
            l=mid;
        } 
        else
            r=mid;
    }
    printf("%.10lf\n",mid );
}