菜雞隻配做水題

思路就很簡單嘛：肯定擴充套件的越靠後邊越好了

 0 0 1 0 1 1 0 0 

假設範圍是3 ，第一個1一定要選上，第2、3個肯定選3啦，越靠後邊就一定能節省更多的點，沒看出來和子問題有什麼聯絡，基本上就是貪心

#include <bits/stdc++.h>
#define ll long long
#define mp make_pair
#define x first
#define y second
using namespace std;
const int maxn = 1005;
int a[maxn];
vector<int
 
> q;
int main()
{
    //freopen("in.txt","r",stdin);
    int n,k;
    scanf("%d %d",&n,&k);
    int cnt = 0;
    q.push_back(0);
    for(int i = 1; i <= n; ++i)
    {
        scanf("%d",a+i);
        if(a[i] == 1){
            q.push_back(i);
            cnt ++;
        }
    }
    int cn = 0
 
;
    int pos = 0;
    int ans = 0;
    while(cn < n)
    {
        int i;
        for(i = pos + 1; i < q.size(); ++i)
        {
            if(cn < q[i]-(k-1)-1)
                break;
        }
        if(i == pos + 1)
        {
            cout << -1 << endl;
            return
 
 0;
        }

        pos = i-1;
        cn = q[pos]+(k-1);
        ans ++;
    }
    cout << ans << endl;
}