The Meeting Place Cannot Be Changed

Question

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.

Example

Input
3
7 1 3
1 2 1
Output
2.00000000000

---

Input
4
5 10 3 2
2 3 2 4
Output
1.400000000000

Code 1

#include 
#include 
#include 
#include 
using namespace std;

double a[60005],b[60005];

int main()
{
    int n;
    int i,j,k;
    double l,r,mid1,mid2,t1,t2;
    cin>>n;
    for(i=0;icin>>a[i];
    for(i=0;icin>>b[i];
    l=0;r=1e9+5;
    while(l+0.0000010;
        mid1=l+(r-l)/2;
        mid2=mid1+(r-mid1)/2;
        for(i=0;ifabs(mid1-a[i])/b[i]);
            t2=max(t2,fabs(mid2-a[i])/b[i]);
        }
        if(t1else
            l=mid1;
    }
    printf("%.12lf\n",min(t1,t2));

    return 0;

Code 2

#include 
#include 
#include 
#include 
#include 
using namespace std;

struct ss
{
    double a,b;
};
bool comp(const ss &a,const ss &b)
{
    return a.aint main()
{
    vectorv;
    ss ab[60005];
    int i,j,k,n;
    double l,r,mid1,mid2;
    double t1,t2;

    cin>>n;
    for(i=0;icin>>ab[i].a;
    for(i=0;icin>>ab[i].b;
    v.clear();
    for(i=0;i0].a;r=1e9+5;
    while(l+0.0000010;
        mid1=l+(r-l)/2;
        mid2=mid1+(r-mid1)/2;
        for(i=0;ifabs(mid1-v[i].a)/v[i].b);
            t2=max(t2,fabs(mid2-v[i].a)/v[i].b);
        }
        if(t1else
            l=mid1;
    }
    printf("%.12lf\n",min(t1,t2));

    return 0;
}

这两个代码都是对集合点进行二分，着最小时间。
实际上这题并不用排序，所以第二题的方法麻烦了一些，但其中对结构体排序的方法还是不错的。
ps:这题我只想强调一点，最后输出的时候 用cout< 或printf("%lf\n",r);会WA。
因为这样输出是默认输出保留小数点6位后输出,即四舍五入到小数点后6位再输出。(这一点，价值三个小时！！！)