开始还以为是尺取。发现行不通。

一看标签二分答案，恍然大悟。

二分一个\(mid\)(实数)，把数列里每个数减去\(mid\)，然后求前缀和，在用单调队列维护\(sum[i-t\text{~}i-s]\)的最小值，用\(sum[i]\)减去它，如果大于等于\(0\)就说明\(mid\)可行。

#include 
    
     #include 
     
      using namespace std;#define INF 2147483647const int MAXN = 100010;const double eps = 1e-6;int n, s, t, head, tail;int a[MAXN];int q[MAXN];inline int read(){    int s = 0, w = 1;    char ch = getchar();    while(ch < '0' || ch > '9'){if(ch == '-')w = -1;ch = getchar();}    while(ch >= '0' && ch <= '9') s = s * 10 + ch - '0',ch = getchar();    return s * w;}int Min = INF, Max = -INF;double l, r, mid, sum[MAXN];inline int check(double mid){    head = tail = 0;    for(int i = 1; i <= n; ++i)       sum[i] = sum[i - 1] + a[i] - mid;    for(int i = s; i <= n; ++i){       int in = i - s;       while(head < tail && sum[in] < sum[q[tail]]) --tail;       q[++tail] = in;       while(head < tail && q[head + 1] < i - t) ++head;       if(sum[i] - sum[q[head + 1]] >= 0) return 1;    }    return 0;}int main(){    n = read();    s = read(); t = read();    for(int i = 1; i <= n; ++i){       a[i] = read();       Min = min(Min, a[i]);       Max = max(Max, a[i]);    }    l = Min; r = Max;    while(r - l > eps){      mid = (l + r) / 2.0;      if(check(mid)) l = mid;      else r = mid;    }    printf("%.3lf\n", l);    return 0;}