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

連結：戳這裡

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.

題意：

給出n個數，找出一個實數x，使得weakness值最小

weakness：表示所有區間內的poorness的最大值

poorness：表示一段區間內的區間和的絕對值

思路：

由於是取絕對值的區間和，所以減去這個x使得值越大不行，越小也不行，肯定存在一個最優值使得它比x左邊和右邊都更優

三分這個極值，每次處理一下就可以了

至於怎麼處理這個絕對值最大問題，我們先-x按區間段最大找一遍，再把陣列去相反數找一次取max

第一次寫三分，好像直接for100次來的更簡單粗暴，毫無精度可言

程式碼：

#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;
}