codeforces 782B 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.

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.
题意:题目已知有n个点，南北走向，那就是y轴了，然后告诉n个点的位置，接下来在告诉你每个点每秒可以上下移动的速度的最大值，问最后最少需要多少时间来汇集所有的点
思路：二分，以最后汇集的点的位置来二分
AC代码：
#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;}这个题目最后有个误差要求，刚开始把精度放得太低错了一次，后面改大就一次过了.