Codeforces Round 1058 (Div. 1)
D1
Inverse Minimum Partition (Easy Version)
Codeforces
2500
原题
← CD2 →
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This is the easy version of the problem. The difference between the versions is that in this version, you are only asked to find $f(a)$. You can hack only if you solved all versions of this problem.

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For some sequence $b$ of $k$ positive integers, the cost of the sequence is defined as follows$^{\text{∗}}$:

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$$\mathtt{cost}(b)=\left\lceil{\frac{b_k}{\min(b_1,b_2,\ldots,b_k)}}\right\rceil$$

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Assume that you partition a sequence $c$ into one or more sequences, so that the sequences become $c$ when concatenated. For example, when the sequence $c$ is $[3,1,4,1,5]$, a few valid ways to partition the sequence are $[[3,1],[4,1,5]]$, $[[3],[1,4,1],[5]]$, $[[3,1,4,1,5]]$. On the other hand, $[[3,1],[1,5]]$ and $[[3,4,1],[1,5]]$ are not valid ways to partition the sequence.

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For a partition of a sequence $c$ of positive integers, the total cost of the partition is defined as the sum of the costs of each sequence in the partition. Then, let us define $f(c)$ as the minimum total cost to partition the sequence $c$.

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Given a sequence $a$ of $n$ positive integers, you must compute the value of $f(a)$.

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$^{\text{∗}}$Given a real value $x$, $\left\lceil{x}\right\rceil$ is defined as the smallest integer no less than $x$. For example, the value of $\left\lceil{3.14}\right\rceil$ is $4$.

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Input

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Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

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The first line of each test case contains a single integer $n$ ($1 \le n \le 4\cdot 10^5$).

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The second line of each test case contains $a_1,a_2,\ldots,a_n$ ($1 \le a_i \le 10^{18}$).

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It is guaranteed that the sum of $n$ over all test cases does not exceed $4\cdot 10^5$.

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Output

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For each test case, output the answer to the problem on a separate line.

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Note

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For the first test case, the partition $[[3,1,4,1],[5]]$ gives a total cost of $1+1=2$.

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For the second test case, the partition $[[9,2,6,5,3],[5,8,9,7,9]]$ gives a total cost of $2+2=4$.

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For the third test case, the partition $[[1],[2,3,4],[5,6,7,8]]$ gives a total cost of $1+2+2=5$.

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For the fourth test case, the partition $[[1],[1\,000\,000\,000\,000\,000\,000]]$ gives a total cost of $1+1=2$.

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样例

Input #1
4
5
3 1 4 1 5
10
9 2 6 5 3 5 8 9 7 9
8
1 2 3 4 5 6 7 8
2
1 1000000000000000000
Output #1
2
4
5
2
InputCopy OutputCopy
题目信息
题号
2159D1
时间限制
4 s
内存限制
1024 MB
标签
二分
暴力
动态规划
计算几何
贪心
数学
双指针

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。