ACM思维题训练集合 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:
丫的，改了一天。 如果b在构造的树的深度最大(左偏或右偏树）和最小（满二叉树)之内就能构成，然后从左偏树开始不断的将低端的点向上移动，知道达到要求。
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
int f[210];
inline void solve()
{
memset(f, 0, sizeof(f));
int n, d, maxd = 0;
scanf("%d %d", &n, &d);
--n;
if (d > n * (n 1) / 2)
{
printf("NOn");
return;
} //1
for (int i = 1;; i)
{
maxd = i;
if (n > (1 << i))
{
d -= i * (1 << i);
f[i] = 1 << i;
n -= 1 << i;
}
else
{
d -= i * n;
f[i] = n;
n -= n;
break;
}
}
if (d < 0)
{
printf("NOn");
return;
}
while (1)
{
if (d == 0)
break;
int p;
for (p = maxd; p >= 1; --p)
if (f[p] > 1)
break;
--d;
--f[p];
f[p 1];
if (p 1 > maxd)
maxd = p 1;
}
printf("YESn");
int p = 1, np = 1, cnt;
for (int i = 1; i <= maxd; i)
{
int t = p;
cnt = 0;
for (int j = 1; j <= f[i]; j)
{
p;
cnt;
if (cnt >= 3)
{
np;
cnt = 1;
}
printf("%d ", np);
}
np = t 1;
}
printf("n");
}
int main()
{
int t;
scanf("%d", &t);
for (int i = 1; i <= t; i)
solve();
return 0;
}
