F - Weakness and Poorness CodeForces

F - Weakness and Poorness CodeForces - 578C

Problem Description

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.

求减去x后的，连续字串和的绝对值的最小值。

题解

x偏大、偏小都会使连续字串的绝对值 deval 偏大，相当于deval 关于 x 的下凸函数，利用三分法查找到最低处的 x ，带入求出连续字串的最大绝对值

复杂度

求连续字串的最大字串的绝对值为 O（n）
三分 O（2log n ）
O(2nlogn ) == 1e7

算法

三分

代码

求连续字串和的绝对值的最大值

分正负讨论
正：逐项求和，如果 sum 出现负值,则将 sum 清零, 继续寻找下一个连续和
负: 每一项都取负值, 则转化为求正值

整体代码

#include<bits\stdc++.h>
using namespace std;
#define maxn 200005
const double eps = 3e-12;
int n;
double a[maxn], l, r, midl, midr, ans, sum = 0;double weak(double x){ // 求连续字串和的绝对值的最大值 // 正数 sum = 0;ans = 0;for(int i = 1; i <= n; i++) {sum += (a[i] - x);ans = max(ans, sum);if(sum <= 0) sum = 0; // 如果sum>0则接下来的字串和将加上sum变得更大，若sum<0则和会减小 }// 负数 sum = 0;for(int i = 1; i <= n; i++) {sum += (x - a[i]); // 全部取负，与正数求法相同 ans = max(ans, sum);if(sum < 0) sum = 0; }return ans;
}int main() {freopen("test.in", "r", stdin);while(scanf("%d", &n) == 1) {l = 10000; r = -10000 ;for(int i = 1; i <= n; i++) {scanf("%lf", &a[i]);if(a[i] < l) l = a[i];if(a[i] > r) r = a[i];}while(r-l >= eps) {midl = l + (r-l)/3.0;midr = r - (r-l)/3.0;if(weak(midl) < weak(midr)) r = midr;else l = midl;}printf("%.15f\n", weak(l));}return 0;
}