CF1491E.Fib-tree
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题目描述
Let Fk denote the k-th term of Fibonacci sequence, defined as below:
- F0=F1=1
- for any integer n≥0, Fn+2=Fn+1+Fn
You are given a tree with n vertices. Recall that a tree is a connected undirected graph without cycles.
We call a tree a Fib-tree, if its number of vertices equals Fk for some k, and at least one of the following conditions holds:
- The tree consists of only 1 vertex;
- You can divide it into two Fib-trees by removing some edge of the tree.
Determine whether the given tree is a Fib-tree or not.
设 Fk 表示斐波那契数列的第 k 项,其定义如下:
- F0=F1=1
- 对任意整数 n≥0,有 Fn+2=Fn+1+Fn
给定一棵包含 n 个顶点的树。回忆一下,树是一种无环的连通无向图。
若一棵树满足以下条件,则称其为斐波那契树(Fib-tree):
- 其顶点数等于某个 k 对应的 Fk;
- 并且满足以下至少一个条件:
- 该树仅含 1 个顶点;
- 存在某条边,将其删除后可将原树划分为两棵斐波那契树。
请判断给定的树是否为斐波那契树。
输入格式
The first line of the input contains a single integer n (1≤n≤2⋅105) — the number of vertices in the tree.
Then n−1 lines follow, each of which contains two integers u and v (1≤u,v≤n, u=v), representing an edge between vertices u and v. It's guaranteed that given edges form a tree.
输入的第一行包含一个整数 n(1≤n≤2⋅105)——树中顶点的数量。
接下来是 n−1 行,每行包含两个整数 u 和 v(1≤u,v≤n,u=v),表示顶点 u 与顶点 v 之间的一条边。保证所给的边构成一棵树。
输出格式
Print "YES" if the given tree is a Fib-tree, or "NO" otherwise.
You can print your answer in any case. For example, if the answer is "YES", then the output "Yes" or "yeS" will also be considered as correct answer.
如果给定的树是一棵 Fib-树,则输出 “YES”,否则输出 “NO”。
您可以以任意大小写形式输出答案。例如,若正确答案是 “YES”,则输出 “Yes” 或 “yeS” 也被视为正确答案。
输入输出样例
输入#1
3 1 2 2 3
输出#1
YES
输入#2
5 1 2 1 3 1 4 1 5
输出#2
NO
输入#3
5 1 3 1 2 4 5 3 4
输出#3
YES
说明/提示
In the first sample, we can cut the edge (1,2), and the tree will be split into 2 trees of sizes 1 and 2 correspondently. Any tree of size 2 is a Fib-tree, as it can be split into 2 trees of size 1.
In the second sample, no matter what edge we cut, the tree will be split into 2 trees of sizes 1 and 4. As 4 isn't Fk for any k, it's not Fib-tree.
In the third sample, here is one possible order of cutting the edges so that all the trees in the process are Fib-trees: (1,3),(1,2),(4,5),(3,4).
在第一个样例中,我们可以切断边 (1,2),此时树将被分割为两棵大小分别为 1 和 2 的子树。任何大小为 2 的树都是 Fib-树,因为它可以进一步被分割为两棵大小均为 1 的子树。
在第二个样例中,无论切断哪条边,树都将被分割为两棵大小分别为 1 和 4 的子树。由于 4 不是任意 k 对应的斐波那契数 Fk,因此它不是 Fib-树。
在第三个样例中,以下是一种可行的删边顺序,使得过程中出现的所有树均为 Fib-树:(1,3), (1,2), (4,5), (3,4)。
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