2019.2.15 t3 平均值

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cmath>
 #include <cctype>
 #include <cstring>
 #include <queue>
 #include <algorithm>
 using namespace std;

 #define res register int
 inline int read()
 {
     int x(),f(); char ch;
     while(!isdigit(ch=getchar())) if(ch=='-') f=-;
     while(isdigit(ch)) x=x*+ch-'',ch=getchar();
     return f*x;
 }

 inline int max(int x,int y){return x>y?x:y;}
 inline double max(double x,double y){return x>y?x:y;}
 inline int min(int x,int y){return x<y?x:y;}

 const int N=+;
 int a[N],n,k,L,R,t;
 int d[N],q[N],q1[N];
 double s[N];//差分数组，前缀和数组 

 inline bool check(double m)
 {
     int l=,r=; //把i看成右端点（此处取得最大s），找s[i]最小的左端点
     double ans(-1e9-);
     for(res i= ; i<=n ; i++)
     {
         if(i-L+<=) continue;
         while(l<=r && s[q[r]]>=s[i-L+]) r--;
         //加入i-L+1(和之前比右端点向右挪了一位)
         q[++r]=i-L+;
         while(i-q[l]+>R && l <= r) l++;//排除非法
         if(l<=r) ans=max(ans,s[i]-s[q[l]]);
     }
     return ans>m*k;
 }

 inline bool judge(double m)
 {
     for(res i= ; i<=n ; i++)
         s[i]=s[i-]+(double)d[i]-m;
     return check(m);
 }
 int main()
 {
     freopen("average.in","r",stdin);
     freopen("average.out","w",stdout);
     int T=read();
     while(T--)
     {
         double maxn(0.0);
         n=read(); k=read(); L=read(); R=read();
         for(res i= ; i<=n ; i++) a[i]=read();
         for(res i= ; i<=n ; i++) d[i]=a[i]-a[i-];
         double l=,r=1e9+;
         int head(),tail(),head1(),tail1();
         double ans(-100000000.0);
         //特殊情况，取得最大值与最小值的两端点之间的距离小于L
         //此处用线段树也可以
         for(res i= ; i<=n ; i++)
         {
             while((head<=tail) && (a[q[tail]]>a[i])) tail--;
             q[++tail]=i;
             while((head1<=tail1) && (a[q1[tail1]]<a[i])) tail1--;
             q1[++tail1]=i;
             while((head<=tail) && i-q[head]+ > L) head++;
             while((head1<=tail1) &&i-q1[head1]+ > L) head1++;
             ans=max(ans,(double)(a[q1[head1]]-a[q[head]]) / (double)(L+k-));
         }
         while(r-l>0.0000001)
         {
             double mid=(l+r)/;
             if(judge(mid)) l=mid,ans=max(ans,l);
             else r=mid;
         }
         //翻转过来，再求一遍，即在刚才的右端点处取得最小s
         reverse(a+,a+n+);
         for(res i= ; i<=n ; i++) d[i]=a[i]-a[i-];
         l=,r=1e9+;
         while(r-l>0.0000001)
         {
             double mid=(l+r)/2.0;
             if(judge(mid)) l=mid,ans=max(ans,l);
             else r=mid;
         }
         printf("%.4lf\n",ans);
     }
     return ;
 }