题解
题目大意 n个灯0关灯1开灯 101则中间的睡不着 问最少关掉多少个灯可以全都能睡着
遇见101则将后面的1的灯泡关掉 这样解决10101的问题 计数输出即可
AC代码
#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e2 + 10;
int a[MAXN];
int main()
{
#ifdef LOCAL
freopen("C:/input.txt", "r", stdin);
#endif
int N;
cin >> N;
for (int i = 1; i <= N; i++)
scanf("%d", &a[i]);
int cnt = 0;
for (int i = 2; i < N; i++)
if (a[i - 1] && !a[i] && a[i + 1]) //101则将后面的关闭
a[i + 1] = 0, cnt++;
cout << cnt << endl;
return 0;
}