The Meeting Place Cannot Be Changed (codeforces 782B)(二分法)

作者: qwq 分类: ACM 发布时间: 2017-07-13 09:04

 

题目:

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.

Example

Input
3
7 1 3
1 2 1
Output
2.000000000000
Input
4
5 10 3 2
2 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.

这个题目的解决办法就是最基本的二分法,二分时间t,从left=0,right=两个点最大距离/最小速度。开始二分。

然后二分t之后对该t进行正义的审判,审判是否可以做到某一个点对所有的人都能到达。

然后逐步逼近最终答案。

这个题比较坑的地方在于该题要求的精度是10^-6 ,但是如果你在代码中改成10^-7精度的话,就会超时。

另一个比较坑的地方在于cout和printf的精度都只有小数点后6位,根本达不到题目要求的精度

必须得另制定printf输出到小数点后10位,才可以AC。

 

#include<iostream>
#include<cstdio>
double a[60001];
double v[60001];
int n;
using namespace std;
int judge(double x){//正义的审判 
	double left[n];
	double right[n];

	for(int i=0;i<n;i++){
		left[i]=a[i]-x*v[i];
		right[i]=a[i]+x*v[i];
					

	}

	double leftmax=left[0],rightmax=right[0];

	for(int i=0;i<n;i++){//判断每个人是否都可以到达某一点(是否有公共区间) 
		leftmax=left[i]>leftmax?left[i]:leftmax;
		rightmax=right[i]>rightmax?rightmax:right[i];
		
	}

	if(leftmax<rightmax){
		
		return 1;
	}else{
		return 0;
	}
	
}
int main(){
	ios::sync_with_stdio(false);
	cin>>n;
		double min=10e9;
	double max=a[0];
	for(int i=0;i<n;i++){
		cin>>a[i];
		min=min>a[i]?a[i]:min;
		max=max>a[i]?max:a[i];
	}
	double lowwer_speed=10e9;
	for(int i=0;i<n;i++){
		cin>>v[i];
		lowwer_speed=lowwer_speed>v[i]?v[i]:lowwer_speed;
	}

	double l=0;
	double r=(max-min)/lowwer_speed;
	double mid=(r-l)/2;
	
	while(r-l>1e-6){
		mid=(r+l)/2;
		if(judge(mid)){
			
			r=mid;
		}else{
			l=mid;
		}
	}
	 printf("%.10llf\n",(r+l)/2);//注意小数点后的位数 
	return 0;
} 

 

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