Minimum Sum LCM - UVa 10791

[title3]
Minimum Sum LCM[/title3]

LCM (Least Common Multiple) of a set of integers is defined 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 or12, 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 find out a set of at least two positive integers whose LCM is N.
As infinite such sequences are possible, you have to pick the sequence whose summation 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 asLCM of 4 and 3 is 12 and 7 is the minimum possible summation.

Input

The input file 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 specified 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

题意：将一个数分解成至少两个数的最小公倍数，求着几个数的和的最小值。

思路：分解素数后，将素数的幂，就是这些数中的一个。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int vis[100010],prime[100010],num;
int main()
{
    int n,n2,t=0,i,j,k,p;
    ll ans;
    for(i=2;i<=100000;i++)
       if(vis[i]==0)
       {
           prime[++num]=i;
           for(j=i*2;j<=100000;j+=i)
              vis[j]=1;
       }
    while(~scanf("%d",&n) && n>0)
    {
        k=0;ans=0;n2=n;
        for(i=1;i<=9592;i++)
           if(n%prime[i]==0)
           {
               k++;p=1;
               while(n%prime[i]==0)
               {
                   n/=prime[i];
                   p*=prime[i];
               }
               ans+=p;
           }
        if(n!=1)
        {
            ans+=n;
            k++;
        }
        ans+=max(0,2-k);
        printf("Case %d: %lld\n",++t,ans);
    }

}