UVA 10791 Minimum Sum LCM （數論）

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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 theLCM of a set of positive integers. For example 12 can be expressed as theLCM 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 find out a set of at least two positive integers whoseLCM 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 and7 is the minimum possible summation.

Input 

The input file contains at most 100 test cases. Each test case consists of a positive integerN ( 1N231 - 1).

Input is terminated by a case where N = 0. This case should not be processed. There can be at most100 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 

 121050

Sample Output 

 Case 1: 7Case 2: 7Case 3: 6

真不想吐槽這道題，前前後後花了2個小時（唯一分解定理的運用）

I64d竟然wa，lld過的，就因為這卡了1個小時。。。。。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>#include<cmath>typedef long long LL;typedef unsigned long long ULL;using namespace std;int main(){   LL n;   int l=0;   while(~scanf("%lld",&n)&&n)   {       LL ans=0;       int num=0;       int m=(int)sqrt(n);       for(int i=2;i<=m;i++)       {           if(n%i==0)           {               num++;               LL sum=1;               while(n%i==0)               {                   n/=i;                   sum*=i;               }               ans+=sum;//               cout<<ans<<endl;           }       }       if(num==0)          ans=(LL)n+1;       else if(n>1||num==1)          ans+=n;       printf("Case %d: %lld\n",++l,ans);   }   return 0;}