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.

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 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.

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.

这题可以用三分，但是在三分判断条件的时候要注意，用循环限定次数比较好，因为double精度不够高。

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<math.h>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<stack>

#include<string>

#include<algorithm>

using namespace std;

#define ll long long

#define inf 99999999

#define maxn 200060

#define eps 1e-10

int n;

double a[maxn],b[maxn],sum1[maxn],sum2[maxn],c[maxn];  

double cal(double x)

{

    int i,j,f;

    double maxnum=0;

    sum1[0]=sum2[0]=0;

    for(i=1;i<=n;i++){

        b[i]=a[i]-x;

        sum1[i]=max(b[i],sum1[i-1]+b[i]);maxnum=max(maxnum,sum1[i]);

        c[i]=x-a[i];

        sum2[i]=max(c[i],sum2[i-1]+c[i]);maxnum=max(maxnum,sum2[i]);

    }

    return maxnum;  

}  

int main()

{

    int m,i,j;

    double l,r,m1,m2,minx,maxx;

    while(scanf("%d",&n)!=EOF)

    {

        for(i=1;i<=n;i++){

            scanf("%lf",&a[i]);

        }

        l=-10005;r=10005;

        for(i=0;i<100;i++){

            double d=(l+r)/2.0;

            m1=d;

            m2=(d+r)/2.0;

            if(cal(m2)-cal(m1)>=0){

                r=m2;

            }

            else l=m1;

        }

        printf("%.9f\n",cal(l));

    }

    return 0;

}