Weakness and Poorness

Weakness and Poorness

time limit per test

2 seconds

memory limit per test

256 megabytes

input

1. 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.

首先，理解为什么要用三分：

　　　　x越小(可为负)，连续序列 最大和 越大;

　　　　x越大，连续序列 最小和 的 绝对值 也越大；

所以，就会随x的递增，形成一个凹函数;

其次，求取连续序列 最大和 与连续序列最小和的 绝对值 ：

　　　　求连续序列最大和:

　　　　设 b[i] = a[i] - x,sum(b[i]) = 连续序列中有 b[i] 的 序列最大和;

　　　　若sum(b[j-1])<0,那么sum(b[j]) = max( b[j],b[j] + sum(b[j-1]) ),sum(b[j])必然等于b[j]，这时sum(b[j])只加 自身 为最优;

　　　　同理，求连续序列最小和，若sum(b[j-1]) > 0;sum(b[j]) = b[j];

至于卡精度的问题，我还是渣渣= =!，大神求教！

复杂度 O(n*log(n))

#include<cstdio>
#include<algorithm>
const double eps = 1e-6;
using namespace std;
int n;
double a[200005];
double b[200005];
double fabs(double x){
    if(x<0)return -x;
    return x;
}
double work(double x){
    for(int i = 0; i < n; i++)
        b[i] = a[i] - x;
    double sum = 0;
    double ans = 0;
    for(int i = 0; i < n; i++){
        sum += b[i];
        if(sum < 0)
            sum = 0;
        ans = max(ans,sum);
    }
    sum = 0;
    for(int i = 0; i < n; i++){
        sum += b[i];
        if(sum > 0)
            sum = 0;
        ans = max(fabs(sum),ans);
    }
    return ans;
}
int main(){
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
        scanf("%lf",&a[i]);
    }
    double l = -10000;
    double r = 10000;
    for(int i = 0; i < 100; i++){
        double lmid = (l+r)/2;
        double rmid = (lmid+r)/2;
        if(work(lmid) > work(rmid))
            l = lmid;
        else r = rmid;
    }
    printf("%.6f\n",work(l));
    return 0;
}