Codeforces Round #320 (Div. 2) [Bayan Thanks-Round]

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.

三分查找：如果遇到凸性或凹形函数时，可以用三分查找求那个凸点或凹点。

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;++i)
#define per(i,a,n) for(int i=n-1;i>=a;--i)
#define mem(a,t) memset(a,t,sizeof(a))
#define pb push_back
#define mp make_pair
#define sz(a) (int)a.size()
#define fi first
#define se second
typedef long long LL;
#define N 200005
int a[N];
int n;
const double eps=1e-12;
double fun(double x)
{
    double tmp=0,ans=0,tmin=0,tmax=0;
    rep(i,0,n){
        tmp+=(a[i]-x);
        ans=max(ans,fabs(tmp-tmin));
        ans=max(ans,fabs(tmp-tmax));
        tmin=min(tmin,tmp);
        tmax=max(tmax,tmp);
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    double l,r;
    scanf("%d",&n);
    l=1e7;
    r=-1e7;
    rep(i,0,n){
        scanf("%d",&a[i]);
        l=min(l,1.*a[i]);
        r=max(r,1.*a[i]);
    }
    double f1,f2,mid1,mid2;
    rep(i,0,500){      
        mid1=l+(r-l)/3;
        mid2=r-(r-l)/3;
        f1=fun(mid1);
        f2=fun(mid2);
        if(f1<f2)
            r=mid2;
        else
            l=mid1;
    }
    cout.precision(10); 
    cout<<fixed<<fun(l)<<endl;
    return 0;
}

ternary search