因為題目裡要求Xi | Xi+1，而Xm又限定為X，所以我們可以想到Xm-1是X除以其某個約數得到的，Xm-1也是一樣。由此我們可以知道“X-factor Chains”是通過不斷乘以X的約數得到的，為了長度最大，所以約數必須是素數。通過記錄有哪些素因數，以及素因子的數量，我們就可以得到鏈的長度。
    而長度最大的鏈的數量則是這些數的排列數，公式為素因子個數之和的階乘/每個素因子個數的階乘。

程式碼（G++）：

#include <iostream>
#include <map>

using namespace std;

map<int
 
, int> res;
map<int, int>::iterator mi;

long long fact(int n)
{
    long long v;

    if(n == 0 || n == 1) return 1;
    else{
        v = n;
        for(int i=2; i<n; i++)
        {
            v *= i;
        }
        return v;
    }
}

int main()
{
    int x, sum;
    long long ans;

    while
 
(cin>>x)
    {
        sum = 0;
        res.clear();
        for(int i=2; i*i<=x; i++)
        {
            while(x%i == 0)
            {
                x /= i;
                ++res[i];
                ++sum;
            }
        }
        if(x != 1)
        {
            ++res[x];
            ++sum;
        }
        ans = fact(sum);
        for
 
(mi=res.begin(); mi!=res.end(); mi++)
        {
            ans /= fact(mi->second);
        }
        cout<<sum<<' '<<ans<<endl;
    }
    return 0;
}

題目：
X-factor Chains
Time Limit: 1000MS Memory Limit: 65536K

Description

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

1 = X0, X1, X2, …, Xm = X

satisfying

Xi < Xi+1 and Xi | Xi+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.

Input

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

Output

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

Sample Input

2
3
4
10
100
Sample Output

1 1
1 1
2 1
2 2
4 6