C. Mortal Kombat Tower Educational Codeforces Round 95 (Rated for Div. 2)

You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n� bosses in this tower, numbered from 11 to n�. The type of the i�-th boss is ai��. If the i�-th boss is easy then its type is ai=0��=0, otherwise this boss is hard and its type is ai=1��=1.

During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.

Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.

Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n� bosses in the given order.

For example: suppose n=8�=8, a=[1,0,1,1,0,1,1,1]�=[1,0,1,1,0,1,1,1]. Then the best course of action is the following:

• your friend kills two first bosses, using one skip point for the first boss;
• you kill the third and the fourth bosses;
• your friend kills the fifth boss;
• you kill the sixth and the seventh bosses;
• your friend kills the last boss, using one skip point, so the tower is completed using two skip points.

You have to answer t� independent test cases.

Input

The first line of the input contains one integer t� (1≤t≤2⋅1041≤�≤2⋅104) — the number of test cases. Then t� test cases follow.

The first line of the test case contains one integer n� (1≤n≤2⋅1051≤�≤2⋅105) — the number of bosses. The second line of the test case contains n� integers a1,a2,…,an�1,�2,…,�� (0≤ai≤10≤��≤1), where ai�� is the type of the i�-th boss.

It is guaranteed that the sum of n� does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑�≤2⋅105).

Output

For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n� bosses in the given order.

Example

input

6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0

output

2
2
2
2
1
0

题目大意：

我和朋友玩游戏，朋友遇到1的怪只能跳过，我可以杀掉1或0的怪，我和朋友能杀一个或两个怪，但不能不杀，问朋友最少跳几次？

思路：

分析一下在什么情况下朋友不得不跳：

1、第一个怪就是1。

2、连着3个怪是1。

 方法：

跑单for查看有多少连着的1。

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"

void solve(){
	ll x,op,cnt=0,sum=0;
	cin >> op >> sum;
	op--;
	while(op--){
		cin >> x;
		if(x)cnt ++;
		else sum = sum+(cnt/3),cnt=0;
	}
	cout << sum+(cnt/3) << endl;
	return ;
}

int main(){
	ll t=1;cin >> t;
	while(t --)solve();
	return 0;
}