Minimum Sum LCM - UVa 10791

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本文探讨了如何将给定的正整数分解为至少两个数的最小公倍数，并寻找元素和最小化的解决方案。通过素因数分解，我们可以找到满足条件的数集，进而计算出其元素和的最小值。

Minimum Sum LCM

$\epsfbox{p10791.eps}$

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.

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);
    }

}