Codeforces Round #403 (Div. 2) B. The Meeting Place Cannot Be Changed

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

题意：开始每个人都在一条数轴上的某个位置上，位置大于等于1都是整数，每个人有个最大移动速度，为在数轴上面哪个点集合，所用的集合时间最短，

求这个时间，集合地点可以不是整数。

思路：如果路径对短那就是求中位数就好了。但速度不一样，这个点一定是中间某个位置，其它位置都会让结果

变大（为什么会这样，自己手动画画图，把路程线段加和就看出来了）。所以结合点跟集合时间是个有极值的函数。

至于题目中的最大速度，为了时间最少，肯定能多快就多快了，直接用最大速度算。

求极值用三分搜索就可以了，可以算是三分的模板题了。

代码如下：

#include <bits/stdc++.h>
const int N = 200010;
using namespace std;
int x
, v
, n;
double check(double local)
{
double ret = 0;
for (int i = 1; i <= n; i++)
{
double a = (double)x[i];
double vt = (double)v[i];
ret = max(ret, fabs(a - local) / vt);
}
return ret;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r" ,stdin);
#endif

while (~scanf("%d", &n))
{
int tmp = 0;
for (int i = 1; i <= n; i++)
{
scanf("%d", x + i);
tmp = max(tmp, x[i]);
}
for (int i = 1; i <= n; i++)
scanf("%d", v + i);
double l = 1, r = tmp;
while (r - l > 1e-6)
{
double mid = (l + r)/2;
double midd = (l + mid)/2;
double dt = check(mid);
double ddt = check(midd);
if (dt > ddt)
r = mid;
else
l = midd;
}
printf("%.8f\n", check(l));
}
return 0;
}