LCM (Least Common Multiple) of a set of integers is de ned as the minimum number, which is a
multiple of all integers of that set. It is interesting to note that any positive integer can be expressed
as the LCM of a set of positive integers. For example 12 can be expressed as the LCM of 1, 12 or
12, 12 or 3, 4 or 4, 6 or 1, 2, 3, 4 etc.
In this problem, you will be given a positive integer
N. You have to nd out a set of at least two positive in-
tegers whose LCM is N. As in nite such sequences are
possible, you have to pick the sequence whose summa-
tion of elements is minimum. We will be quite happy
if you just print the summation of the elements of this
set. So, for N = 12, you should print 4+3 = 7 as
LCM of 4 and 3 is 12 and 7 is the minimum possible
summation.
Input
The input le contains at most 100 test cases. Each
test case consists of a positive integer N (1 N
231 �� 1).
Input is terminated by a case where N = 0. This
case should not be processed. There can be at most
100 test cases.
Output
Output of each test case should consist of a line starting with `Case #: ’ where # is the test case number. It should be followed by the summation as speci ed in the problem statement. Look at the
output for sample input for details.
Sample Input
12
10
5
0
Sample Output
Case 1: 7
Case 2: 7
Case 3: 6
【分析】
唯一分解定理。
设分解式为 n=a1^p1+a2^p2+…+am^pm(^表示几次方，不是异或)，那么不难发现把每个 ai^pi 作为一个单独的整数时解最优。
这道题真是long long天坑…WA死了
【代码】
