瞎搞题啊。找出1 1 0 0这种序列，然后存起来，这种情况下最好的选择是1的个数除以这段的总和。然后从前向后扫一遍，变扫边进行合并。每次合并，合并的是他的前驱。这样到最后从t-1找出的那条链就是最后满足条件的数的大小。
Room and Moor
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 307 Accepted Submission(s): 90
Problem Description PM Room defines a sequence A = {A
1, A
2,..., A
N}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B
1, B
2,... , 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
Source 2014 Multi-University Training Contest 6
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