Minimal Power of Prime【数学（找规律）】

Minimal Power of Prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5331 Accepted Submission(s): 328

Problem Description
You are given a positive integer n > 1. Consider all the different prime divisors of n. Each of them is included in the expansion n into prime factors in some degree. Required to find among the indicators of these powers is minimal.

Input
The first line of the input file is given a positive integer T ≤ 50000, number of positive integers n in the file. In the next T line sets these numbers themselves. It is guaranteed that each of them does not exceed 10^18.

Output
For each positive integer n from an input file output in a separate line a minimum degree of occurrence of a prime in the decomposition of n into simple factors.

Sample Input
5
2
12
108
36
65536

Sample Output
1
1
2
2
16

题目大意：
首先知道两点之间的权值例如点$a$和点$b$之间的权值等于$a\&b$，有$T$组测试数据，对于每组测试数据，输入一个数$n$，代表有$n$个点，此时需建立一棵树，使得树的权值最小。需先输出最小权值为多少，再输出从点$2$到点$n$每个点连接的点是哪个。

解题思路：
由于需要输出最小权值的树，就需使得边的权值尽可能小，由于点$a$和点$b$之间的权值等于$a\&b$，我们可以发现对于任意一个点，我们都可以找到另一个点与之取与后使得边的权值为$0$，例如当$n$等于$6$时，对于点$2$,其二进制第一个$0$出现在第一位，所以与点$(1<<0)$取与后权值为$0$，对于点$3$，由于$n==6$其二进制为$3$位，对$3$的二进制补零后找出其第一次出现的$0$的位置，在第$3$位，将其点$(1<<2)$相连，其边权值也就变为了$0$，以此类推，最后即可得到最小权值树，且可得到最小的字典树序列。
ps.当$n==(1<<k)-1$时，此时第$n$个点找不到第一个$0$，即找不到一个点与其相连使得边权值为$0$，此时只能将其与点$1$相连。因此我们可以发现，最后的树的最小权值要么为$0$、要么为$1$。

代码：

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