【codeforces 782B】The Meeting Place Cannot Be Changed（二分）

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
4000
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.

ps：二分法枚举时间，第i人在time时间内可到达的区间范围为（x[i]-v[i]*time,x[i]+v[i]*time），然后判断所有人都能到达的范围是否有交集，如果有交集，说明他们都能在time时间内集合

代码：

#include<stdio.h>

#define maxn 60000+10
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
int x[maxn],v[maxn];
int n;

int canmeet(double time)
{
double left,right,le,ri;
for(int i=1; i<=n; i++)
{
if(i==1)
left=x[i]-v[i]*time,right=x[i]+v[i]*time;
else
{
le=x[i]-v[i]*time,ri=x[i]+v[i]*time;
if(le>right||ri<left)
return 0;
left=max(left,le);//维护区间最大的左边值
right=min(ri,right);//维护区间最小的右边值
}
}
return 1;
}

int main()
{
scanf("%d",&n);
for(int i=1; i<=n; i++) scanf("%d",x+i);
for(int i=1; i<=n; i++) scanf("%d",v+i);
double l=0,r=1e9;
while(r-l>1e-6)
{
double mid=(l+r)/2;
if(canmeet(mid))//能在mid时间内能集合
r=mid;
else
l=mid;
}
printf("%.12lf\n",r);
return 0;
}