題意：找到一個連續區間，區間的長度至少大於f，現在要求這個區間的平均值最大。

題解：

二分找答案。

每次對於2分的mid值， 都把原來的區間減去mid， 然後找到一長度至少為f的區間， 他們的區間和>=0。

程式碼：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<queue>
 5 #include<map>
 6 #include<iostream>
 7 #include<cstring>
 8
 
 using namespace std;
 9 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
10 #define LL long long
11 #define ULL unsigned LL
12 #define fi first
13 #define se second
14 #define pb push_back
15 #define lson l,m,rt<<1
16 #define rson m+1,r,rt<<1|1
17 #define lch(x) tr[x].son[0]
18
 
 #define rch(x) tr[x].son[1]
19 #define max3(a,b,c) max(a,max(b,c))
20 #define min3(a,b,c) min(a,min(b,c))
21 typedef pair<int,int> pll;
22 const int inf = 0x3f3f3f3f;
23 const LL INF = 0x3f3f3f3f3f3f3f3f;
24 const LL mod =  (int)1e9+7;
25 const int N = 1e5 + 100;
26 int n, f;
27 int a[N];
28 LL suf[N];
 
29 bool check(int mid){
30     for(int i = 1; i <= n; ++i)
31         suf[i] = suf[i-1] + a[i] - mid;
32     LL minval = 0;
33     for(int i = f; i <= n; ++i){
34         minval = min(minval, suf[i-f]);
35         if(suf[i] - minval >= 0) return true;
36     }
37     return false;
38 }
39 int main(){
40     scanf("%d%d", &n, &f);
41     for(int i = 1; i <= n; ++i)
42         scanf("%d", &a[i]), a[i] *= 1000;
43     int l = 0, r = inf;
44     while(l <= r){
45         int m = l+r >> 1;
46         if(check(m)) l = m + 1;
47         else r = m - 1;
48     }
49     printf("%d\n", l-1);
50     return 0;
51 }