Codeforces Round #320 (Div. 1) C. Weakness and Poorness

Codeforces Round #320 (Div. 1) C. Weakness and Poorness"/>

Codeforces Round #320 (Div. 1) C. Weakness and Poorness

C. 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+DP

不过 要注意精度

 

 

#include <iostream>
#include<stdio.h>
using namespace std;
int a[300000],n;
double _fabs(double val)
{return (val>0?val:-val);
}
double check(double val)
{double _max,sum;_max=0;sum=0;for(int i=1;i<=n;i++){sum=(sum>0?sum:0)+a[i]-val;_max=max(_max,sum);}sum=0;for(int i=1;i<=n;i++){sum=(sum<0?sum:0)+a[i]-val;_max=max(_max,-sum);}return _max;
}
int main()
{double left,right,mid1,mid2;while(scanf("%d",&n)!=EOF){for(int i=1;i<=n;i++)scanf("%d",&a[i]);left=-100000;right=100000;while(_fabs(left-right)>1e-11){mid1=(2*left+right)/3.0;mid2=(left+2*right)/3.0;if (check(mid1)<check(mid2))right=mid2;elseleft=mid1;}printf("%.12lf\n",check((left+right)/2.0));}return 0;
}
/*
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
*/

　　

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