POJ3421(质因数分解)

X-factor Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6501		Accepted: 2023

Description

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.

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
思路:将x进行质因数分解，X-factor Chains中（第因一项永为1，所以不计）第i项就是将质因数进行排列后前i项的积。所以length为质因数的个数。numbeu为质因数全排列的种数。相同质因数之间的排列要去重。

#include <iostream>
#include <map>
using namespace std;
typedef unsigned long long ull;
map<int,int> prime_factor(int n)//质因数分解
{
    map<int,int> res;
    for(int i=;i*i<=n;i++)
    {
        while(n%i==)
        {
            res[i]++;
            n/=i;
        }
    }
    if(n!=)
    {
        res[n]++;
    }
    return res;
}
ull fact(int n)
{
    ull res=;
    for(int i=;i<=n;i++)
    {
        res*=i;
    }
    return res;
}
int x;
int main()
{
    while(cin>>x)
    {
        map<int,int> res=prime_factor(x);
        int len=;
        for(map<int,int>::const_iterator it=res.begin();it!=res.end();it++)
        {
            len+=it->second;
        }
        ull cnt=fact(len);
        for(map<int,int>::const_iterator it=res.begin();it!=res.end();it++)
        {
            cnt/=fact(it->second);//去重
        }
        cout<<len<<" "<<cnt<<endl;
    }
    return ;
}