题目链接:点击查看
题意:n个点,构建1-n-1的k叉树,问每个k叉树,满足子代权值小于父亲的有多少个
题解:查询的次数大约为n+n/2+n/3+...n/n 差不多是n*log(n) 的复杂度,建个主席树,直接查询即可
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
struct node{
int l,r;
int val;
}tree[N*22];
int root[N],cnt;
int n,q;
int a[N],b[N],len;
int build(int l,int r)
{
int cur=++cnt;
tree[cur].val=0;
if(l==r) return cur;
int mid=(r+l)>>1;
tree[cur].l=build(l,mid);
tree[cur].r=build(mid+1,r);
return cur;
}
int update(int l,int r,int pos,int pre)
{
int cur=++cnt;
tree[cur]=tree[pre];
tree[cur].val++;
if(l==r) return cur;
int mid=(r+l)>>1;
if(pos<=mid)tree[cur].l=update(l,mid,pos,tree[pre].l);
else tree[cur].r=update(mid+1,r,pos,tree[pre].r);
return cur;
}
int query(int l,int r,int pos,int x,int y)
{
if(r<=pos)
{
return tree[y].val-tree[x].val;
}
int res=0;
int mid=(r+l)>>1;
res+=query(l,mid,pos,tree[x].l,tree[y].l);
if(pos>mid)res+=query(mid+1,r,pos,tree[x].r,tree[y].r);
return res;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&b[i]),a[i]=b[i];
sort(b+1,b+1+n);
len=unique(b+1,b+1+n)-(b+1);
cnt=0;
root[0]=build(1,len);
int pos;
for(int i=1;i<=n;i++)
{
pos=lower_bound(b+1,b+1+len,a[i])-b;
root[i]=update(1,len,pos,root[i-1]);
}
int ans;
int l,r;
for(int i=1;i<n;i++)
{
ans=0;
for(int j=1;i*(j-1)+2<=n;j++)
{
pos=lower_bound(b+1,b+1+len,a[j])-b;
l=i*(j-1)+2,r=min(n,i*j+1);
pos--;
if(pos)
{
ans+=query(1,len,pos,root[l-1],root[r]);
}
}
printf("%d%c",ans," \n"[i==n-1]);
}
return 0;
}