UVa 10791 Minimum Sum LCM (number theory & element factorization)

10791-minimum Sum LCM

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php? Option=com_onlinejudge&itemid=8&category=467&page=show_problem&problem=17 32

LCM (least Common multiple) of a set of integers is defined as the minimum number, which are a multiple of all int Egers of that set. It is interesting to the positive integer can expressed as the LCM of a set of positive integers. For Example can expressed as the LCM of 1, or 3, 4 or 4, 6 or 1, 2, 3, 4 etc.

In the problem, you'll be given a positive integer N. You are have to find out a set of in least two positive integers whose LCM is N. As infinite such sequences are possible, you are have to pick the sequence whose the summation of-is elements. We are quite happy if you are just print the summation of the elements of this set. So, for N = should print4+3 = 7 as LCM of 4 and 3 are 7 is the minimum possible summation.

Input

The input file contains at most test cases. Each test case consists of a positive integer n (1n2 to 1).

The Input is terminated by a case where N = 0. This case should is processed. There can be at most test cases.

Output

Output of each test case should consist's 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 in the output for sample input for details.

Sample Input

5
0

Sample Output

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

Thinking: Element factorization

Complete code:

/*0.022s*/
  
#include <cstdio>
#include <cmath>
  
int main (void)
{
    int n, ncase = 0;
    int maxi, temp, F, I;
    Long ans;
    while (scanf ("%d", &n), N)
    {
        maxi = (int) sqrt (n);
        Ans = 0;
        f = 0;
        for (i = 2; I <= maxi i++)
            if (n% i = = 0)
            {
                f++;
                temp = 1;
                while (n% i = = 0)
                {
                    temp *= i;
                    n/= i;
                }
                Ans + temp;
            }
  
        Supplement
        if (f = = 0)
            ans = (long long) n + 1;
        else if (n > 1 | | f = = 1)
            ans + = n;
        printf ("Case%d:%lld\n", ++ncase, ans);
    }
    return 0;
}

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