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下载 Olymp Trade 应用程序手机

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PPTP is most common VPN that almost all device have it in their setting menu. PPTP is simple VPN that you don't need install app to run it. You can go to our totorial in VPN JANTIT if you want more how to use PPTP Free PPTP Server location around the world. PPTP is simple to use. Support for Android, Windows, Mikrotik, Linux. Active up to 7 days with unlimited 下载 Olymp Trade 应用程序手机 bandwidth.

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下载 Olymp Trade 应用程序手机

AJAO TOOSIN 交易者

2021-08-29 20:16 发表于 尼日利亚

我最近被这个应用程序骗了,存入了 52000 奈拉,他们声称在接下来的 24 小时内将我的钱翻倍,但24 小时后什么也没收到,我打电话给客户服务,他们都拉黑了我,请远离他们。

I got scammed from this app of recent,I deposited 52000 naira to this app claiming to double my money for me in the next 24hours,after 24hrs nothing was received and I even call their customer care they all blocked me,please run away from them

Codeforces Round #765 (Div. 2)B. Elementary Particles

~晚风微凉~ 于 2022-01-13 01:04:36 发布 189 收藏 1

Martians are actively engaged in interplanetary trade. Olymp City, the Martian city known for its spaceport, has become a place where goods from all the corners of our Galaxy come. To deliver even more freight from faraway planets, Martians need fast spaceships.

A group of scientists conducts experiments to build a fast engine for the new spaceship. In the current experiment, there are n elementary particles, the i-th of them has type ai.

Denote a subsegment of the particle sequence (a1,a2,…,an) as a sequence (al,al+1,…,ar) for some left bound l and right bound r (1≤l≤r≤n). For instance, the sequence (1 4 2 8 5 7) 下载 Olymp Trade 应用程序手机 for l=2 and r=4 has the sequence (4 2 8) as a subsegment. Two subsegments are considered different if at least one bound of those subsegments differs.

Note that the 下载 Olymp Trade 应用程序手机 subsegments can be equal as sequences but still considered different. For example, consider the sequence (1 1 1 1 1) and two of its subsegments: one with l=1 and r=3 and another with l=2 and r=4. Both subsegments are equal to (1 1 1), but still considered different, 下载 Olymp Trade 应用程序手机 下载 Olymp Trade 应用程序手机 as their left and right bounds differ.

The scientists want to conduct a reaction to get two different subsegments of the same length. Denote this length k. The resulting pair of subsegments must be harmonious, i. e. for some i (1≤i≤k) it must be true that the types of particles on the i-th position are the same for these two subsegments. For example, the pair (1 7 3) and (4 7 8) is harmonious, as both subsegments have 7 on the second position. The pair (1 2 3) and (3 1 2) is not harmonious.

The longer are harmonious subsegments, the more chances for the scientists to design a fast engine. So, they asked you to calculate the maximal possible length of harmonious pair made of different subsegments.

Input
The first line contains an integer t (1≤t≤100) — the number of test cases. The following are descriptions of the test cases.

The first line contains an integer n (2≤n≤150000) — the amount of elementary particles in the sequence.

The second line contains n integers ai (1≤ai≤150000) — types of elementary particles.

It is guaranteed that the sum of n over all test cases does not exceed 3⋅105.

Output
For each test, print a single integer, maximal possible length of harmonious pair made of different subsegments. If 下载 Olymp Trade 应用程序手机 such pair does not exist, print −1 instead.

Example
inputCopy
4
7
3 1 5 2 1 3 下载 Olymp Trade 应用程序手机 4
6
1 1 1 1 1 1
6
1 4 2 8 5 7
2
15 15
outputCopy
4
5
-1
1
Note
The first test case is shown on the picture below:

As you can see from it, you may choose the subsegments (2 1 3 4) and (3 1 5 2), which are a harmonious pair. Their length is equal to 4, so the answer is 4.

In the second test case, you need to take two subsegments: one with l=1 and r=5, and one with l=2 and r=6. It’s not hard to observe that these segments are a harmonious pair and considered different even though they are both equal to (1 1 1 1 1).

In the third test case, you cannot make 下载 Olymp Trade 应用程序手机 a harmonious pair, so the answer is −1.
好久没写cf了,希望之后能够上分
题意:
就是要你找到能够让两个长度相同(但不完全相同的)子串有某个同一位置的数字相同,求那个最大的子串长度