Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] E 三分+连续子序列的和的绝对值的最大值

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, ..., a_n.

Determine a real number x such that the weakness of the sequence a₁ - x, a₂ - x, ..., a_n - 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, ..., a_n (|a_i| ≤ 10 000).

Output

Output a real number denoting the minimum possible weakness of a₁ - x, a₂ - x, ..., a_n - 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：   给你一段序列a₁, a₂, ..., a_n   

            a₁ - x, a₂ - x, ..., a_n - x.    对于每一个x 都有ans=连续子序列的和的绝对值的最大值

            输出min（ans）

题解：   贪心求出连续子序列的和的绝对值的最大值 o（n）处理

           三分x （x为实数存在负数）  求min（ans）

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int n;
double a[200005];
double fun(double x)
{
    double sum1=0.0,sum2=0.0;
    double max1=0.0,min1=100005.0;
    for(int i=1;i<=n;i++)
    {
        if((sum1+a[i]-x)<0)
            sum1=0;
        else
            sum1=sum1+a[i]-x;
        if((sum2+a[i]-x)>0)
            sum2=0;
        else
            sum2=sum2+a[i]-x;
        max1=max(max1,sum1);
        min1=min(min1,sum2);
    }
    return max(abs(max1),abs(min1));
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    scanf("%lf",&a[i]);
    double l=-1e9,r=1e9,m1=0.0,m2=0.0;
    for(int i=0;i<600;i++)
    {
        m1=l+(r-l)/3.0;
        m2=r-(r-l)/3.0;
        if(fun(m1)<fun(m2))
            r=m2;
        else
            l=m1;
    }
   printf("%f
",fun(l));
    return 0;
}