A10016. Weakness and Poorness
编程题
普及/提高-
知识点
题目描述
You are given a sequence of n integers $a_{1},a_{2},...,a_{n}$ .
Determine a real number $x$ such that the weakness of the sequence $a_{1}-x,a_{2}-x,...,a_{n}-x$ 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.
Determine a real number $x$ such that the weakness of the sequence $a_{1}-x,a_{2}-x,...,a_{n}-x$ 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.
输入格式
The first line contains one integer $n$ ( $1<=n<=200000$ ), the length of a sequence.
The second line contains $n$ integers $a_{1},a_{2},...,a_{n}$ ( $|a_{i}|<=10000$ ).
The second line contains $n$ integers $a_{1},a_{2},...,a_{n}$ ( $|a_{i}|<=10000$ ).
输出格式
Output a real number denoting the minimum possible weakness of $a_{1}-x,a_{2}-x,...,a_{n}-x$ . Your answer will be considered correct if its relative or absolute error doesn't exceed $10^{-6}$ .
输入输出样例
输入 #1
3 1 2 3
输出 #1
1.000000000000000
输入 #2
4 1 2 3 4
输出 #2
2.000000000000000
输入 #3
10 1 10 2 9 3 8 4 7 5 6
输出 #3
4.500000000000000
说明/提示
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.
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.