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
input
4
5 10 3 2
2 3 2 4
output
1.400000000000
1 #include<cstdio>
2 struct type{
3 int x;
4 int v;
5 }p[60000+5];
6 int n;
7 bool check(double time)
8 {
9 //遍历每个人在time个单位时间后能走到的位置，不断更新重叠的区间[l,r]，只要到最后这个区间一人不为空，就return true
10 double l=p[1].x-time*p[1].v;
11 double r=p[1].x+time*p[1].v;
12 for(int i=2;i<=n;i++){
13 if(p[i].x-time*p[i].v > l) l=p[i].x-time*p[i].v;
14 if(p[i].x+time*p[i].v < r) r=p[i].x+time*p[i].v;
15 if(l>r) return false;
16 }
17 return true;
18 }
19 int main()
20 {
21 scanf("%d",&n);
22 for(int i=1;i<=n;i++) scanf("%d",&p[i].x);
23 for(int i=1;i<=n;i++) scanf("%d",&p[i].v);
24 double st=0,ed=1000000000;
25 while(ed-st>1e-7){
26 double mid=st+(ed-st)/2;
27 if(check(mid)) ed=mid;
28 else st=mid;
29 }
30 printf("%.12lf\n",ed);
31 }
