B. The Meeting Place Cannot Be Changed

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本文解析了一道算法题——如何计算使n个朋友在一条直线上某个点相聚所需的最短时间。通过二分查找的方法求解，并讨论了精度控制问题。

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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 $\text{[math]}$ 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.

一个很水的二分，但因精度问题T3也很气。。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define eps 1e-8
#define PI 3.141592653589793
#define bs 1000000007
#define bsize 256
#define MEM(a) memset(a,0,sizeof(a))
typedef long long ll;
using namespace std;
int x[60050],v[60050];
int n;
int check(double &t)
{
	int i;
	double le,ri;
	le=x[0]-v[0]*t;
	ri=x[0]+v[0]*t;
	for(i=1;i<n;i++)
	{
		le=max(le,x[i]-v[i]*t);
		ri=min(ri,x[i]+v[i]*t);
		if(ri<le)
		return 0;
	}
	return 1;
}
int main()
{
	double l,r,mid,ans;
	int i;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	scanf("%d",&x[i]);	
	for(i=0;i<n;i++)
	scanf("%d",&v[i]);
	l=0,r=1e9;
	while(r-l>1e-7)
	{
		mid=(l+r)/2;
		if(check(mid))
		{
			r=mid;
			ans=mid;
		}
		else
		{
			l=mid;
		}
	}
	printf("%.10lf\n",ans);
	return 0;
 }

精度1e-8时，上面的代码会t，但下面这个基本一样的代码，区别就是把check放到主函数里，这个就可以过，在编译器运行test3都是无限循环（试了codeblocks和dev），可是交了就能过，很迷。。。。。。

3
1 1000000000 2
1 2 1000000000

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define eps 1e-8
#define PI 3.141592653589793
#define bs 1000000007
#define bsize 256
#define MEM(a) memset(a,0,sizeof(a))
typedef long long ll;
using namespace std;
int x[60050],v[60050];
int n;
int main()
{
	double l,r,mid,ans,m,le,ri;
	int i;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	scanf("%d",&x[i]);	
	for(i=0;i<n;i++)
	scanf("%d",&v[i]);
	l=0,r=1e9;
	while(r-l>1e-8)
	{
		mid=(l+r)/2;
		int flog=1;
		le=x[0]-v[0]*mid;
		ri=x[0]+v[0]*mid;
		for(i=1;i<n;i++)
		{
			le=max(le,x[i]-mid*v[i]);
			ri=min(ri,x[i]+mid*v[i]);
			if(ri<le)
			{
				flog=0;
				break;
			}
		}
		if(flog)
		{
			r=mid;
			ans=mid;
	//		printf("%.9lf %.9lf\n",r,l);
		}
		else
		{
			l=mid;
		}
	}
	printf("%.10lf\n",ans);
	return 0;
 }