这道题用到了唯一分解定理，例如，我们将12通过唯一分解定理分解，得到2的平方*3，2的平方换成4，这样3加4就等于了7.
所以这道题的解题关键就是将输入值通过唯一分解定理分解，然后相加即可。
不过有特例，输入是1时，答案为1+1=2；以及输入是质数时(即只有一种因子)需要加一
另外看清本题的数据大小，不要溢出
不打表版：
1 #include<iostream>
2 #include<cstring>
3 #include<cmath>
4
5 typedef unsigned long long ull;
6
7
8 using namespace std;
9
10 ull lcm(ull a);
11
12 int main()
13 {
14
15 ios::sync_with_stdio(false);
16 cin.tie(0);
17 cout.tie(0);
18
19 ull num;
20 int times = 1;
21
22 while (cin >> num)
23 {
24 if (num == 0)
25 break;
26 ull end = lcm(num);
27 cout << "Case "<<times++<<": "<< end << endl;
28 }
29
30 return 0;
31 }
32
33 ull lcm(ull a)
34 {
35 if (a == 1)
36 return 2;
37
38 ull m = (ull)sqrt(a+0.5);
39 int j = 0;
40 ull sum = 0;
41
42 for (ull i = 2; i <= m; i++)
43 {
44 if (a%i == 0)
45 {
46 ull k=1;
47
48 while (a%i == 0)
49 {
50 k *= i;
51 a /= i;
52 }
53 j++;
54 sum += k;
55
56 }
57 }
58
59 if (a > 1 || j == 0)
60 {
61 sum += a;
62 j++;
63 }
64
65 if (j == 1)
66 sum++;
67
68 return sum;
69
70
71 }
打素数表版：
1 #include<iostream>
2 #include<cstring>
3 #include<cmath>
4
5 typedef unsigned long long ull;
6 const int maxn =46340+20;//2的31次方-1加0.5开平方，然后加20
7 int primes[maxn];
8 bool p[maxn];
9 int tol = 0;
10
11 using namespace std;
12
13 ull lcm(ull a);
14 void find_primes(ull n);
15
16 int main()
17 {
18
19 ios::sync_with_stdio(false);
20 cin.tie(0);
21 cout.tie(0);
22
23 ull num;
24 int times = 1;
25 find_primes(maxn);
26
27 while (cin >> num)
28 {
29 if (num == 0)
30 break;
31 ull end = lcm(num);
32 cout << "Case "<<times++<<": "<< end << endl;
33 }
34
35 return 0;
36 }
37
38 ull lcm(ull a)
39 {
40 if (a == 1)
41 return 2;
42
43 ull m = (ull)sqrt(a+0.5);
44 int j = 0;
45 int t = 0;
46 ull sum = 0;
47
48 for (ull i = primes[t]; i <= m; i=primes[++t])
49 {
50 if (a%i == 0)
51 {
52 ull k=1;
53
54 while (a%i == 0)
55 {
56 k *= i;
57 a /= i;
58 }
59 j++;
60 sum += k;
61
62 }
63 }
64
65 if (a > 1 || j == 0)
66 {
67 sum += a;
68 j++;
69 }
70
71 if (j == 1)
72 sum++;
73
74 return sum;
75
76
77 }
78
79 void find_primes(ull n)
80 {
81 memset(p, false, sizeof(p));
82
83 for (int i = 2; i <= n; i++)
84 {
85 if (!p[i])
86 {
87 primes[tol++] = i;
88 for (int j = i * i; j <= n; j += i)
89 p[j] = true;
90 }
91 }
92 }