poj3421 X-factor Chains

题意：

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：

一开始想到dp。令dp[i][j]表示长度为i，以j结尾的链的个数，于是dp[i+1][k] += dp[i][j] (j为k的因子)，然而复杂度高，并不会优化。

后来发现要想链最长，只能从1开始，每次乘上这个数的某个质因子才行。于是就变成了分解质因子+排列组合的问题。

实现：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
typedef long long ll;

ll fac[21];
int x;

void init()
{
    fac[0] = 1;
    for (ll i = 1; i <= 20; i++)
        fac[i] = fac[i - 1] * i;
}

map<int, int> prime_factor(int n)
{
    map<int, int> res;
    for (int i = 2; i * i <= n; i++)
    {
        while (n % i == 0)
        {
            res[i]++;
            n /= i;
        }
    }
    if (n != 1)
        res[n] = 1;
    return res;
}

int main()
{
    init();
    while (scanf("%d", &x) != EOF)
    {
        map<int, int> f = prime_factor(x);
        map<int, int>::iterator it;
        ll res = 1;
        int cnt = 0;
        for (it = f.begin(); it != f.end(); it++)
        {
            res *= fac[it->second];
            cnt += it->second;
        }
        printf("%d %lld
", cnt, fac[cnt] / res);
    }
    return 0;
}