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

31 2 3

output

1.000000000000000

input

41 2 3 4

output

2.000000000000000

input

101 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;}