2020 BIT冬训-二分三分快速幂矩阵 F - Weakness and Poorness CodeForces - 578C

Problem Description

You are given a sequence of n integers a₁,?a₂,?...,?a_n.

Determine a real number x such that the weakness of the sequence a₁?-?x,?a₂?-?x,?...,?a_n?-?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,?...,?a_n (|a_i|?≤?10?000).

Output

Output a real number denoting the minimum possible weakness of a₁?-?x,?a₂?-?x,?...,?a_n?-?x. Your answer will be considered correct if its relative or absolute error doesn‘t exceed 10^?-?6.

Examples

3
1 2 3

1.000000000000000

4
1 2 3 4

2.000000000000000

10
1 10 2 9 3 8 4 7 5 6

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.

这题的思路是三分。

由于减去的x过大或过小，weakness都会过大。当x为极小值点时才会最小。即下凸函数。

三分查找最小值即可。

weakness就是数列的最大的连续子串的和的绝对值。因此分最大绝对值的原数是正数和负数讨论下即可。

精度为2e-12,3e-12。（低了会wa，高了会tle）不是很懂（应该是答案误差需要小于1e-6。n个精度相加对weakness的影响为1e-7）

#include<cstdio>
#include<algorithm>
using namespace std;
int n;
double a[200005],b[200005],r=-10005,l=10005,midl,midr,sum,ans;
double weakn(double m){
    ans=0;
    sum=0;
    for(int i=0;i<n;i++){
        sum+=a[i]-m;
        if(sum<0)
            sum=0;
        ans=max(ans,sum);
    }
    sum=0;
    for(int i=0;i<n;i++){
        sum+=m-a[i];
        if(sum<0)
            sum=0;
        ans=max(ans,sum);
    }
    return ans;
}
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lf",&a[i]);
        if(a[i]>r)
            r=a[i];
        if(a[i]<l)
            l=a[i];
    }
    while(r-l>2e-12){
        midl=(l+r)/2;
        midr=(midl+r)/2;
        if(weakn(midl)<weakn(midr))
            r=midr;
        else
            l=midl;
    }
    printf("%lf",weakn(l));
}

2020 BIT冬训-二分三分快速幂矩阵 F - Weakness and Poorness CodeForces - 578C

原文：https://www.cnblogs.com/mikku39/p/14406934.html