E. Construct the Binary Tree------------------------思维(构造二叉树)

You are given two integers n and d. You need to construct a rooted binary tree consisting of n vertices with a root at the vertex 1 and the sum of depths of all vertices equals to d.

A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a vertex v is the last different from v vertex on the path from the root to the vertex v. The depth of the vertex v is the length of the path from the root to the vertex v. Children of vertex v are all vertices for which v is the parent. The binary tree is such a tree that no vertex has more than 2 children.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤1000) — the number of test cases.

The only line of each test case contains two integers n and d (2≤n,d≤5000) — the number of vertices in the tree and the required sum of depths of all vertices.

It is guaranteed that the sum of n and the sum of d both does not exceed 5000 (∑n≤5000,∑d≤5000).

Output
For each test case, print the answer.

If it is impossible to construct such a tree, print “NO” (without quotes) in the first line. Otherwise, print “{YES}” in the first line. Then print n−1 integers p2,p3,…,pn in the second line, where pi is the parent of the vertex i. Note that the sequence of parents you print should describe some binary tree.

Example
inputCopy
3
5 7
10 19
10 18
outputCopy
YES
1 2 1 3
YES
1 2 3 3 9 9 2 1 6
NO
Note
Pictures corresponding to the first and the second test cases of the example:

题意:
给你n个点，深度之和为d，问你是否能构建出这颗二叉树，如果可以输出2~n每个节点的父亲节点。否 输出NO
解析:
深度之和最大 :所有的节点全部在一条链上
深度之和最小:一颗满二叉树，

我们先创建一颗满二叉树，然后逐步向一条深链转换，直至转换到满足条件即可。详细看代码(有注释)

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10000;
int vis[N],dep[N],fa[N],node[N];
int t,maxd,n,d;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&n,&d);//转化成完全二叉树 
		maxd=0;memset(vis,0,sizeof vis);fa[1]=0;dep[1]=0;
		for(int i=2;i<=n;i++)
		{
			fa[i]=i/2;//记录i的父亲节点 
			dep[i]=dep[i/2]+1;//记录i的深度 
			d-=dep[i];
			maxd=max(maxd,dep[i]);//记录最大深度 
		}
		if(d<0) //他比最小可能还要小，说明肯定无法构造 
		{
			cout<<"NO"<<endl;
			continue;
		}
		int now=n;//找到最深链 
		while(now)
		{
			vis[now]=1;
			node[dep[now]]=now;//链上深度为i指向了j
			now=fa[now]; 
		}
		for(int i=n;i>=1;i--)//完全二叉树向最深链转换 
		{
			if(vis[i]) continue;
			int pre=maxd;
			while(dep[i]<=pre&&d) //更新i点至最深链node[dep[i]] 
			{
				fa[i]=node[dep[i]];
				dep[i]+=1;//因为放在node[dep[i]]下面所以深度++ 
				if(dep[i]>maxd)//更新最大深度 
				{
					maxd=dep[i];
					node[maxd]=i;
				}
				d--;
			}
		}
		if(d>0)//链是最长的不存在剩余 
		{
			cout<<"NO"<<endl;
			continue;
		}
		cout<<"YES"<<endl;
		for(int i=2;i<=n;i++) cout<<fa[i]<<" ";
		cout<<endl;
	}
}