B. 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 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.

思路：二分；

二分时间，然后每次检查时求出每个点所能到达的左右的距离，然后求线段的最大重叠。复杂度（n*log^2（n））

或者二分每次检查时求出每个点所能到达的左右的距离，然后维护最小的右端r，和最大的左端l，如果r > l成立复杂度n*log(n);

 1 #pragma comment(linker, "/STACK:102400000,102400000")
 2 #include<stdio.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<string.h>
 6 #include<stdlib.h>
 7 #include<queue>
 8 #include<map>
 9 #include<math.h>
10 #include<vector>
11 const int N = 1e6+6;
12 using namespace std;
13 typedef long long LL;
14 double x[70000];
15 double v[70000];
16 typedef struct node
17 {
18     double val;
19     int k;
20 } ss;
21 bool cmp(node p,node q)
22 {
23     if(p.val == q.val)
24         return p.k > q.k;
25     else return p.val < q.val;
26 }
27 ss ask[70000*2];
28 bool check(double c,int n);
29 int main(void)
30 {
31     int n;
32     scanf("%d",&n);
33     for(int i = 1; i <= n; i++)
34         scanf("%lf",&x[i]);
35     for(int i = 1; i <= n; i++)
36         scanf("%lf",&v[i]);
37     int t = 100;
38     double ll = 0,rr = 1e10;
39     double acc = -1;
40     while(t)
41     {
42         double mid = (ll+rr)/2;
43         if(check(mid,n))
44         {
45             acc = mid;
46             rr = mid;
47             //printf("%lf
",mid);
48         }
49         else ll = mid;
50         t--;
51     }
52     printf("%.10f
",acc);
53     return 0;
54 }
55 bool check(double c,int n)
56 {
57     int cn = 0;
58     for(int i = 1; i <= n; i++)
59     {
60         ask[cn].val = max((double)0,x[i] - c*v[i]);
61         ask[cn].k = 1;
62         cn++;
63         ask[cn].val = x[i] + c*v[i];
64         ask[cn].k = -1;
65         cn++;
66     }
67     int sum = 0;
68     sort(ask,ask+cn,cmp);
69     for(int i = 0; i < cn; i++)
70     {
71         sum += ask[i].k;
72         if(sum == n)
73             return true;
74     }
75     return false;
76 }

View Code

 1 #pragma comment(linker, "/STACK:102400000,102400000")
 2 #include<stdio.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<string.h>
 6 #include<stdlib.h>
 7 #include<queue>
 8 #include<map>
 9 #include<math.h>
10 #include<vector>
11 const int N = 1e6+6;
12 using namespace std;
13 typedef long long LL;
14 double x[70000];
15 double v[70000];
16 typedef struct node
17 {
18     double val;
19     int k;
20 } ss;
21 bool cmp(node p,node q)
22 {
23     if(p.val == q.val)
24         return p.k > q.k;
25     else return p.val < q.val;
26 }
27 ss ask[70000*2];
28 bool check(double c,int n);
29 int main(void)
30 {
31     int n;
32     scanf("%d",&n);
33     for(int i = 1; i <= n; i++)
34         scanf("%lf",&x[i]);
35     for(int i = 1; i <= n; i++)
36         scanf("%lf",&v[i]);
37     int t = 100;
38     double ll = 0,rr = 1e10;
39     double acc = -1;
40     while(t)
41     {
42         double mid = (ll+rr)/2;
43         if(check(mid,n))
44         {
45             acc = mid;
46             rr = mid;
47             //printf("%lf
",mid);
48         }
49         else ll = mid;
50         t--;
51     }
52     printf("%.10f
",acc);
53     return 0;
54 }
55 bool check(double c,int n)
56 {
57     int cn = 0;
58     double maxx = 1e10;
59     double minn = 0;
60     for(int i = 1; i <= n; i++)
61     {
62         minn = max(minn,x[i] - c*v[i]);
63         maxx = min(maxx,x[i] + c*v[i]);
64     }
65     if(maxx >= minn)return true;
66     return false;
67 }

n*log(n)