Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] E. Weakness and Poorness 三分

E. Weakness and Poorness

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence of n integers a1, a2, ..., an.

Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.

The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.

The poorness of a segment is defined as the absolute value of sum of the elements of segment.

Input

The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.

The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).

Output

Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.

Examples

input

3
1 2 3

output

1.000000000000000

input

4
1 2 3 4

output

2.000000000000000

input

10
1 10 2 9 3 8 4 7 5 6

output

4.500000000000000

Note

For the first case, the optimal value of x is 2 so the sequence becomes  - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.

For the second sample the optimal value of x is 2.5 so the sequence becomes  - 1.5,  - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.

题意：找出一个x，使得a[i]-x的这段连续最大子序列和绝对值最小；

思路：三分，check的时候来回扫一遍

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+,M=1e6+,inf=;
const ll INF=1e18+,mod=;
double a[N],v[N];
int n;
double equ(double x)
{
    double ans=;
    double sum=;
    for(int i=;i<=n;i++)
    {
        sum+=(a[i]-x);
        if(sum<)
            sum=;
        ans=max(ans,sum);
    }
    sum=;
    for(int i=;i<=n;i++)
    {
        sum-=(a[i]-x);
        if(sum<)
            sum=;
        ans=max(ans,sum);
    }
    return ans;
}
double ternarySearch(double l,double r)
{
    for(int i=;i<=;i++)
    {
        double lll=(*l+r)/;
        double rr=(l+*r)/;
        double ans1=equ(lll);
        double ans2=equ(rr);
        if(ans1>ans2)
            l=lll;
        else
            r=rr;
    }
    return l;
}
int main()
{
    scanf("%d",&n);
    for(int i=;i<=n;i++)
        scanf("%lf",&a[i]);
    double ans=ternarySearch(-,);
    printf("%f\n",equ(ans));
    return ;
}