NetEase C + + face test--place Street lamp

Little Q is designing a street lamp placement scheme for a road of length n.

To make the problem easier, small Q treats the road as n squares, where it needs to be illuminated. Indicates that the barrier lattice that does not need to be illuminated is denoted by ' X '.

Small q now to set some street lights on the road, for the position of the street lamp, this lamp can illuminate pos-1, POS, pos + 1 of these three positions.

Little Q hopes to place as few streetlights as possible to illuminate all '. ' Area, I hope you can help him calculate the minimum number of lights required.

Input Description:

The first line of the input contains a positive integer t (1 <= T <= 1000), indicating the number of test cases
Next every two lines of a test data, the first line is a positive integer n (1 <= n <= 1000), which represents the length of the road.
The second line, a string s, represents the construction of a road, containing only '. ' and ' X '.

Output Description:

For each test case, the output of a positive integer indicates the minimum number of lights required.

Input Example 1:

23.x.11 ... Xx.... Xx

Output Example 1:

13

#include <bits/stdc++.h>using namespace Std;typedef long long ll;string s;const int maxn = 1010;int Bits[maxn];int m Ain () {    int T;    scanf ("%d", &t);    int Len;    while (t--)    {        scanf ("%d", &len);        cin>>s;        memset (bits,0,sizeof (bits));        for (int i=0;i<len;i++)        {            if (s[i]== '. ') bits[i]=0;            else bits[i] = 1;        }        int ans = 0;        for (int i=0;i<len;i++)        {            if (bits[i]) continue;            else            {                ans++;                Bits[i] = 1;                BITS[I+1] = 1;                BITS[I+2] =1;            }        }       for (int i=0;i<len;i++)         //   printf ("%d", bits[i]);        cout<<endl;        printf ("%d\n", ans);}    }

　　

NetEase C + + face test--place Street lamp