578C. Weakness and Poorness(Codeforces Round #320)

Weakness and Poorness(Codeforces Round #320)"/>

578C. Weakness and Poorness(Codeforces Round #320)

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.

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.

题目大意：

      给出一段序列，序列中的每个数都减x，求减去x后的最大连续子序列和和最小连续子序列和的绝对值最小

值。

解题思路：

     很容易理解，对所有的x，减去x后的最大连续子序列和最小连续子序列是呈现单谷函数的，可以用三分来做，

至于最大连续子序列和最小连续子序列的求法，只要2遍最大连续和就可以了，第1遍后将每个数取反，这样第2

遍也是一个最大连续和。

     还有一种方法是二分，随着x的递增，序列的最大连续子序列和肯定在减小，因为每个数都减小了，其最小连

续子序列和的绝对值肯定是增大趋势，都是单调的，于是将最大连续子序列的和和最小连续子序列和的绝对值分开记录，就可二分，会有点麻烦。

 ps：被精度坑了，1e-11会wa，1e-12会T，三分100次就过了，通过这次学到了浮点三分根据数据通过控制三分次数更好。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const double eps=1e-11;
const int maxn=200000+100;
int a[maxn];
double l[maxn];
int n;
double cal(double x)
{double sum=0;for(int i=0;i<n;i++)l[i]=a[i]*1.0-x;double cur=0;for(int i=0;i<n;i++){cur=cur+l[i];if(cur<0)cur=0;sum=max(sum,cur);l[i]=-l[i];}cur=0;for(int i=0;i<n;i++){cur=cur+l[i];if(cur<0)cur=0;sum=max(sum,cur);}return sum;
}
int main()
{scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&a[i]);}double l=-1e4,r=1e4,m1,m2,v1,v2;int t=100;while(t--){m1=l+(r-l)/3;m2=r-(r-l)/3;v1=cal(m1);v2=cal(m2);if(v1>v2)l=m1;else r=m2;}printf("%.15f\n",cal(l));return 0;
}