Given a positive integer X, an X-factor chain of length m is a sequence of integers,

1 = X₀, X₁, X₂, …, X_m = X

satisfying

X_i < X_i+1 and X_i | X_i+1 where a | b means a perfectly divides into b.

Now we are interested in the maximum length of X-factor chains and the number of chains of such length.

The input consists of several test cases. Each contains a positive integer X (X ≤ 2²⁰).

For each test case, output the maximum length and the number of such X-factors chains.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <vector>
 5 #include <algorithm>
 6 #include <map>
 7 using namespace std;
 8 typedef long long ll;
 9
10 ll fac[21];
11 int x;
12
13 void init()
14 {
15     fac[0] = 1;
16     for (ll i = 1; i <= 20; i++)
17         fac[i] = fac[i - 1] * i;
18 }
19
20 map<int, int> prime_factor(int n)
21 {
22     map<int, int> res;
23     for (int i = 2; i * i <= n; i++)
24     {
25         while (n % i == 0)
26         {
27             res[i]++;
28             n /= i;
29         }
30     }
31     if (n != 1)
32         res[n] = 1;
33     return res;
34 }
35
36 int main()
37 {
38     init();
39     while (scanf("%d", &x) != EOF)
40     {
41         map<int, int> f = prime_factor(x);
42         map<int, int>::iterator it;
43         ll res = 1;
44         int cnt = 0;
45         for (it = f.begin(); it != f.end(); it++)
46         {
47             res *= fac[it->second];
48             cnt += it->second;
49         }
50         printf("%d %lld\n", cnt, fac[cnt] / res);
51     }
52     return 0;
53 }