Codeforces 578C Weakness and Poorness

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我是靠谱客的博主故意秋天，最近开发中收集的这篇文章主要介绍Codeforces 578C Weakness and Poorness，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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.

解题思路：所求的值在[-10000,10000]为下凹函数的性质，因此我们可以采用三分法求解。注意精度问题，一开始自己写的三分各种WA，后来改了个小地方过了。。。。。。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 200010;
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const ll  INF = 0x3f3f3f3f3f3f3f3fLL;
double arr[maxn];
int n;

double cal(double x) {
	double t1 = 0, t2 = 0, maxsum = -INF, minsum = INF; 
	for(int i = 1; i <= n; ++i) {
		if(t1 >= 0) {
			t1 += arr[i] - x;
		} else {
			t1 = arr[i] - x;
		}
		if(t2 <= 0) {
			t2 += arr[i] - x;
		} else {
			t2 = arr[i] - x;
		}
		maxsum = max(maxsum, t1);
		minsum = min(minsum, t2);
	}
	return max(abs(maxsum), abs(minsum));
}
int main() {
	
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		scanf("%lf", &arr[i]);
	}
	double l = -10005.0, r = 10005.0;
	int lim = 100;
	while(lim--) {
		double ll = l + (r - l) / 3.0;
		double rr = r - (r - l) / 3.0;
		if(cal(ll) > cal(rr)) {
			l = ll;
		} else {
			r = rr;
		}
	}
	printf("%.9lfn", cal(l));
	return 0;
}