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

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）

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 using namespace std;
 6 int n;
 7 double a[200005];
 8 double fun(double x)
 9 {
10     double sum1=0.0,sum2=0.0;
11     double max1=0.0,min1=100005.0;
12     for(int i=1;i<=n;i++)
13     {
14         if((sum1+a[i]-x)<0)
15             sum1=0;
16         else
17             sum1=sum1+a[i]-x;
18         if((sum2+a[i]-x)>0)
19             sum2=0;
20         else
21             sum2=sum2+a[i]-x;
22         max1=max(max1,sum1);
23         min1=min(min1,sum2);
24     }
25     return max(abs(max1),abs(min1));
26 }
27 int main()
28 {
29     scanf("%d",&n);
30     for(int i=1;i<=n;i++)
31     scanf("%lf",&a[i]);
32     double l=-1e9,r=1e9,m1=0.0,m2=0.0;
33     for(int i=0;i<600;i++)
34     {
35         m1=l+(r-l)/3.0;
36         m2=r-(r-l)/3.0;
37         if(fun(m1)<fun(m2))
38             r=m2;
39         else
40             l=m1;
41     }
42    printf("%f
",fun(l));
43     return 0;
44 }