(a[i+1]+a[i+2]+......+a[i+x])/x<=k
(a[i+1]-k)+(a[i+2]-k)+......(a[i+x]-k)<=0
b[i+1]=a[i+1]-k
s[i]=b[1]+b[2]+......+b[i]
统计:j<i s[j]>s[i]
用树状数组维护 O(nlogn)
这种想法以前接触过，忘记......
