codeforces578C. Weakness and Poorness

Weakness and Poorness"/>

codeforces578C. 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.

这题可以用三分，但是在三分判断条件的时候要注意，用循环限定次数比较好，因为double精度不够高。

#include<iostream>  
#include<stdio.h>  
#include<stdlib.h>  
#include<string.h>  
#include<math.h>  
#include<vector>  
#include<map>  
#include<set>  
#include<queue>  
#include<stack>  
#include<string>  
#include<algorithm>  
using namespace std;  
#define ll long long  
#define inf 99999999  
#define maxn 200060  
#define eps 1e-10  
int n;  
double a[maxn],b[maxn],sum1[maxn],sum2[maxn],c[maxn];  double cal(double x)  
{  int i,j,f;  double maxnum=0;  sum1[0]=sum2[0]=0;  for(i=1;i<=n;i++){  b[i]=a[i]-x;  sum1[i]=max(b[i],sum1[i-1]+b[i]);maxnum=max(maxnum,sum1[i]);  c[i]=x-a[i];  sum2[i]=max(c[i],sum2[i-1]+c[i]);maxnum=max(maxnum,sum2[i]);  }  return maxnum;  }  int main()  
{  int m,i,j;  double l,r,m1,m2,minx,maxx;  while(scanf("%d",&n)!=EOF)  {  for(i=1;i<=n;i++){  scanf("%lf",&a[i]);  }  l=-10005;r=10005;  for(i=0;i<100;i++){  double d=(l+r)/2.0;m1=d;m2=(d+r)/2.0;if(cal(m2)-cal(m1)>=0){  r=m2;  }  else l=m1;  }  printf("%.9f\n",cal(l));  }  return 0;  
}