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.
InputThe 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.
OutputPrint 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 NoteIn 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：原来没有做过这一类的题目，在网上看了一下题解才明白，这个题目是用二分法来做，而二分的东西正是时间，我们搜索到一个时间使所有的人都能移动到同一个点，而小于这个时间有的人就无法移动到这一个点，这也就是要保证所有的点在时间t内能移动到的点都有交集。明白了这一点以后，就不难求解了。
#include<iostream> #include<algorithm> #include<cstring> #include <string> #include <set> #include <queue> #include <cmath> #include <iomanip> #include <vector> #include <cstdio> #define MAXN 60010 #define INF 10000000 using namespace std; int v[MAXN]; int p[MAXN]; int n; bool judge(double time) { double left = p[0] - v[0] * time, right = p[0] + v[0] * time; for (int i = 1; i < n; ++i) { double l = p[i] - v[i] * time; double r = p[i] + v[i] * time; if (r < left || l > right) return false; right = min(r, right); left = max(left, l); } return true; } int main() { while (~scanf("%d", &n)) { for (int i = 0; i < n; ++i) scanf("%d", &p[i]); for (int i = 0; i < n; ++i) scanf("%d", &v[i]); double left = 0, right = 1e9; while (right - left > 1e-6) { double mid = (right + left) / 2; if (judge(mid)) right = mid; else left = mid; } printf("%.12f", right); } return 0; }