用一个栈维护b的值，每次把一个数放到栈顶。看栈首的值是不是大于这个数，如果大于的话将栈顶2个元素合并，b的值就是这两个栈顶元素的平均值。。。

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Problem Description

PM Room defines a sequence A = {A₁, A₂,..., A_N}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B₁, B₂,... , B_N} of the same length,
which satisfies that:

Input

The input consists of multiple test cases. The number of test cases T(T<=100) occurs in the first line of input.

For each test case:

The first line contains a single integer N (1<=N<=100000), which denotes the length of A and B.

The second line consists of N integers, where the ith denotes A_i.

Output

Output the minimal f (A, B) when B is optimal and round it to 6 decimals.

Sample Input

4
9
1 1 1 1 1 0 0 1 1
9
1 1 0 0 1 1 1 1 1
4
0 0 1 1
4
0 1 1 1

Sample Output

1.428571
1.000000
0.000000
0.000000

Author

BUPT

Source

2014 Multi-University Training Contest 6

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <cmath>

using namespace std;

const double eps=1e-8;

double a[110000],b[110000];

typedef pair<double,double> pDD;

int n;
stack<pDD> STACK;

int main()
{
	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d",&n);
		while(STACK.size()) STACK.pop();
		for(int i=0;i<n;i++)
		{
			scanf("%lf",a+i);
			pDD A=pDD(a[i],1);
			while(STACK.size()&&A.first+eps<STACK.top().first)
			{
				pDD B=STACK.top(); STACK.pop();
				double Sec=A.second+B.second;
				double Fst=(A.first*A.second+B.first*B.second)/Sec;
				A.first=Fst; A.second=Sec;
			}
			STACK.push(A);
		}

		int now=n-1;
		while(STACK.size())
		{
			pDD u=STACK.top(); STACK.pop();
			int sz=u.second;
			for(int i=now,j=0;i>=0&&j<sz;i--,j++)
			{
				b[now--]=u.first;
			}
		}

		double ans=0;
		for(int i=0;i<n;i++)
		{
			ans+=(a[i]-b[i])*(a[i]-b[i]);
		}
		printf("%.6lf\n",ans);
	}
	return 0;
}

HDOJ 4923 Room and Moor