题面#

Consider the following cube $D$ where numbers $x$ and $7-x$ lie on opposite sides:

A sequence $b$ of integers from $1$ to $6$ is called a dice roll sequence if it satisfies the following condition:

• All pairs of adjacent elements lie on adjacent $^{\text{∗}}$ sides of the cube.

For example, $[1,4,2]$ is a dice roll sequence, while $[3,4,6,3]$ is not because $3$ and $4$ are not on adjacent sides of the dice. Additionally, $[2,2,4]$ is not a dice roll sequence because $2$ and $2$ are on the same (not adjacent) side of the dice.

Given a sequence $a$ of $n$ integers from $1$ to $6$, you can perform the following operation any number of times (possibly zero).

• Select an index $1 \le i \le n$ and an integer $1 \le x \le 6$. Then, change the value of $a_i$ to $x$.

Please determine the minimum number of operations required to make $a$ a dice roll sequence.

$^{\text{∗}}$Two sides of the cube $S$ and $T$ are called adjacent if they share exactly one edge of the cube. Do note that this condition implies $S \neq T$ as well.

Input

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.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1 \le a_i \le 6$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output

For each test case, output the minimum number of operations required to make $a$ a dice roll sequence.

思路#

由于要保证相邻的两个元素不能在骰子上是“相对”的，所以我们只需要遍历一遍这个序列，然后找相邻的两个元素是否相对，利用

来进行判断

代码#

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#include <bits/stdc++.h>
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using namespace std;
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int main() {
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    int t;
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    cin >> t;
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    while (t --) {
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        int n;
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        cin >> n;
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        vector<int> a(n);
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        for (int i = 0; i < n; i ++) cin >> a[i];
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        int count = 0;
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        for (int i = 1; i < n; i ++) {
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            if (a[i] + a[i - 1] == 7 || a[i] == a[i - 1]) {
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                count ++;
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                a[i] = 0;
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            }
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        }
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        cout << count << endl;
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    }
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}