ACM刷题之Codeforces ————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.
比赛时被卡住的一道题。
赛后看了题解，发现是二分。
研究了一下，然后做了一下，做出来了
首先假定最大时间是从0到最长距离的那个时间（速度1）
然后进行二分。
只要往左的最大和往右的最小有交集，则放大L，没有则缩小R。
下面是ac代码 （注意精度）
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<map>#include<set>#include<queue>#include<string>#include<iostream>using namespace std;#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const int N=2e5+7;const double cha=0.0000001;double v[N];double x[N];double l,r,r2,maxn=0.0,mids;int n;bool check(double mi){	double maxx=0.0,minxx=1000000000;	for(int i=1;i<=n;i++)	{		maxx=max(maxx,x[i]-v[i]*mi);		minxx=min(minxx,x[i]+v[i]*mi);			}	if(maxx<minxx)			return false;				return true;}int main(){	//freopen("f:/input.txt", "r", stdin);	scanf("%d",&n);	for(int i=1;i<=n;i++)	{		scanf("%lf",&x[i]);		maxn=max(x[i],maxn);	}	for(int i=1;i<=n;i++)	{		scanf("%lf",&v[i]);	}	l=0;	r=r2=maxn;	while(r-l>cha)	{		mids=(r+l)/2;		if(check(mids))		{			l=mids+cha;		}else{			r=mids-cha;		}	}	printf("%lf/n",l-cha);	}