Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] E 三分

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E. Weakness and Poorness

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

题意：

给出n个数，找出一个实数x，使得weakness值最小

weakness：表示所有区间内的poorness的最大值

poorness：表示一段区间内的区间和的绝对值

思路：

由于是取绝对值的区间和，所以减去这个x使得值越大不行，越小也不行，肯定存在一个最优值使得它比x左边和右边都更优

三分这个极值，每次处理一下就可以了

至于怎么处理这个绝对值最大问题，我们先-x按区间段最大找一遍，再把数组去相反数找一次取max

第一次写三分，好像直接for100次来的更简单粗暴，毫无精度可言

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n;
double a[200100],b[200100];
const double eps=1e-12;
double getmax(){
double ans=0,sum=0;
for(int i=1;i<=n;i++){
sum=sum+b[i];
if(sum<0) sum=0;
ans=max(ans,sum);
}
return ans;
}
double calc(double x){
for(int i=1;i<=n;i++) b[i]=a[i]-x;
double ans=getmax();
for(int i=1;i<=n;i++) b[i]=-b[i];
ans=max(ans,getmax());
return ans;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%lf",&a[i]);
double l=-10000.0,r=10000.0,mid;
for(int i=0;i<100;i++){
double m1=l+(r-l)/3;
double m2=r-(r-l)/3;
if(calc(m1)<calc(m2)) r=m2;
else l=m1;
}
printf("%.12f\n",calc(l));
return 0;
}