天天看点

【Disturbed People】【CodeForces - 1077B】(思维水题)

题目:

There is a house with nn flats situated on the main street of Berlatov. Vova is watching this house every night. The house can be represented as an array of nninteger numbers a1,a2,…,ana1,a2,…,an, where ai=1ai=1 if in the ii-th flat the light is on and ai=0ai=0 otherwise.

Vova thinks that people in the ii-th flats are disturbed and cannot sleep if and only if 1<i<n1<i<n and ai−1=ai+1=1ai−1=ai+1=1 and ai=0ai=0.

Vova is concerned by the following question: what is the minimum number kk such that if people from exactly kk pairwise distinct flats will turn off the lights then nobody will be disturbed? Your task is to find this number kk.

Input

The first line of the input contains one integer nn (3≤n≤1003≤n≤100) — the number of flats in the house.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (ai∈{0,1}ai∈{0,1}), where aiaiis the state of light in the ii-th flat.

Output

Print only one integer — the minimum number kk such that if people from exactly kkpairwise distinct flats will turn off the light then nobody will be disturbed.

Examples

Input

10
1 1 0 1 1 0 1 0 1 0
      

Output

2
      

Input

5
1 1 0 0 0
      

Output

Input

4
1 1 1 1
      

Output

Note

In the first example people from flats 22 and 77 or 44 and 77 can turn off the light and nobody will be disturbed. It can be shown that there is no better answer in this example.

There are no disturbed people in second and third examples.

解题报告:数据出水了,所以直接暴力就可以ac 。如果一个不开灯的人两边都是开灯的话,他无法正常入眠,问至少关闭需要关闭多少灯才可以!

ac代码:

#include<bits/stdc++.h>
using namespace std;

int num[155];
int main()
{
	int n;
	int cnt=0;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%d",&num[i]);
	for(int i=1;i<n-1;i++)
	{
		if(num[i-1]==1&&num[i]==0&&num[i+1]==1)
		{
			cnt++;
			num[i+1]=0;
		}
	}
	printf("%d\n",cnt);
}