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.

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.

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.

37 1 31 2 1output
2.000000000000input
45 10 3 22 3 2 4output
1.400000000000Note
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.
題意:題目已知有n個(gè)點(diǎn)，南北走向，那就是y軸了，然后告訴n個(gè)點(diǎn)的位置，接下來(lái)在告訴你每個(gè)點(diǎn)每秒可以上下移動(dòng)的速度的最大值，問(wèn)最后最少需要多少時(shí)間來(lái)匯集所有的點(diǎn)
思路：二分，以最后匯集的點(diǎn)的位置來(lái)二分
AC代碼：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <map>#include <vector>#include <set>using namespace std;const int MAX_N = 600005;double pos[MAX_N];double speed[MAX_N];int n;int main() {    double mi=0xfffffffff,mx=-1,mid;    cin >> n;    for(int i =1; i <= n; i++) {        cin >> pos[i];        mx = max(mx,pos[i]);        mi = min(mi,pos[i]);    }    for(int i =1; i <= n; i++) {        cin >> speed[i];    }    double shang  = 0,xia = 0;    while(fabs(mi-mx) >= 0.000001) {        mid = (mi+mx)/2;        shang = 0;        xia = 0;        for(int i = 1; i <= n; i++) {            if(pos[i] > mid) {                shang = max(shang,(pos[i]-mid)/speed[i]);            }            if(pos[i] < mid) {                xia = max(xia,(mid-pos[i])/speed[i]);            }        }        if(shang > xia) {            mi = mid;        }        if(shang == xia) {            break;        }        if(shang < xia) {            mx = mid;        }    }    printf("%.12f/n",(shang+xia)/2);    return 0;}這個(gè)題目最后有個(gè)誤差要求，剛開(kāi)始把精度放得太低錯(cuò)了一次，后面改大就一次過(guò)了.