codeforces#320(div2) E. Weakness and Poorness 三分

codeforces#320(div2) 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.

Sample test(s)

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.

显然，随着x从-INF到INF的过程中，所求值是先减后增的，求最小值所以直接三分法。

虽然D题FST了，但是却学到了E题的三分法。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)

using namespace std;

typedef long long ll;
const ll INF=(1LL<<33);
const double EPS=0.00000001;
const int maxn=1000100;

int n;
double a[maxn],b[maxn],c[maxn];

double MaxLong(int op)
{
    REP(i,1,n) c[i]=b[i]*op;
    double now=0,res=c[1];
    REP(i,1,n){
        if(now+c[i]>=0) now+=c[i];
        else now=0;
        res=max(res,now);
    }
    return  res;
}

double f(double x)
{
    REP(i,1,n) b[i]=a[i]-x;
    //REP(i,1,n) cout<<b[i]<<" ";cout<<endl;
    return max(MaxLong(1),MaxLong(-1));
}

double bin3(double l,double r)
{
    while(l<r){
        double m=(l+r)/2;
        double mm=(m+r)/2;
        //printf("%.2f %.2f %.2f %.2f\n",l,r,m,mm);
        double fm=f(m),fmm=f(mm);
        //printf("%.2f %.2f\n",fm,fmm);
        if(fabs(fm-fmm)<EPS) return fm;
        if(fm>fmm+EPS) l=m;
        else r=mm;
    }
}

int main()
{
    freopen("in.txt","r",stdin);
    while(cin>>n){
        REP(i,1,n) scanf("%lf",&a[i]);
        //printf("%.10f\n",f(5.375));
        //REP(i,1,n) cout<<b[i]<<" ";cout<<endl;
        double ans=bin3(-1.0*INF,INF*1.0);
        printf("%.6f\n",ans);
    }
    return 0;
}