B. The Meeting Place Cannot Be Changed----二分答案

B. The Meeting Place Cannot Be Changed

time limit per test
5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

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.

题目链接：http://codeforces.com/contest/782/problem/B

这场cf打得我，真是，哎，这个题的意思是说有一条直线，上面站着n个人，每个人站在ai的位置上，它可以往两边移动，速度为vi，问你所有人汇集到一个点至少需要多长时间。

一开始我想到了二分答案，然后卡在了怎么判断是否汇集到了一个点上，然后时间复杂福也觉得hold不住，然后就gg了，瞎写了一个代码，到第15组怎么都过不去，然后wa到死，cf的b我都打不出来了，还是菜啊。

我们二分时间，然后判断所有人可以走到的左边的最大值，右边的最小值，判断有没有交集即可。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
double a[70000];
double b[70000];
int n;
bool judge(double time){
double left=0,right=0;
for(int i=0;i<n;i++){
if(i==0){
left=a[i]-b[i]*time;
right=a[i]+b[i]*time;
}
else{
double ll=a[i]-b[i]*time;
double rr=a[i]+b[i]*time;
if(ll>right||rr<left){//没有交集
return false;
}
left=max(ll,left);
right=min(rr,right);
}
}
return true;
}
int main(){
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%lf",&a[i]);
}
for(int i=0;i<n;i++){
scanf("%lf",&b[i]);
}
double left=0,right=1e9;
while(right-left>1e-6){
double mid=(left+right)/2;
if(judge(mid)){
right=mid;
}
else{
left=mid;
}
}
printf("%.12lf\n",right);//好想是1e-6就可以
return 0;
}