CF-Round#403 Div2 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

37 1 31 2 1

output

2.000000000000

input

45 10 3 22 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：用二分法

代码如下：

#include<stdio.h>#include<string.h>#define maxn 60000+2int a[maxn];int b[maxn];int n;double max(double a, double b){ return a > b ? a : b; }double min(double a, double b){ return a > b ? b : a; }bool judge(double time){double left = 0, right = 0;for (int i = 0; i < n; i++){if (i == 0){left = a[i] - b[i] * time;right = a[i] + b[i] * time;}else{double l = a[i] - b[i] * time;double r = a[i] + b[i] * time;if (r<left || l>right)//l<r<left（小）<right（大）或者left（小）<right（大）<l<r,即没有交集{return false;}//取交集left = max(left, l);right = min(right, r);}}return true;}int main(){memset(a, 0, sizeof(a));while (scanf("%d", &n) != EOF){for (int i = 0; i < n;i++){scanf("%d",&a[i]);}for (int i = 0; i < n; i++){scanf("%d", &b[i]);}double left = 0, right = 1e9;while (right - left>1e-6){double mid = (left + right) / 2;if (judge(mid)) right = mid;else left = mid;}printf("%.12lf\n", right);}return 0;}

参考：http://blog.csdn.net/bless924295/article/details/60959933