POJ：3421-X-factor Chains（因式分解）（全排列）

X-factor Chains

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 7986 Accepted: 2546

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

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解题心得：

1. 题意就是给你一个数，要求你输出将这个数分解成因式相乘，并且后面一个因子至少是前面一个因子的2倍，问最长的因式相乘链有多长，有几条最长的因式相乘链。
2. 其实就是将一个数分解成若干个素数相乘，因为后面分解出来的素数一定是在前面分解之后分解的，所以自然就大2倍以上。然后问有多少条链，其实就是问将所有因子的全排列有多少种，但是要除去重复因子的全排列。
3. n个不同的数的全排列就是n！，除去重复的就是n！除以每个重复因子数的阶乘。

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#include <map>
#include <algorithm>
#include <stdio.h>
#include <vector>
using namespace std;
typedef long long ll;

map <int,int> prim_factor(ll n2) {
    map<int,int> maps;
    int N = n2;
    for(int i=2;i*i<=N;i++)  {
        while(n2 % i == 0) {
            maps[i]++;
            n2 /= i;
        }
    }
    if(n2 > 1)
        maps[n2]++;
    return maps;
};

ll mult(int k) {
    ll ans = 1;
    for(int i=2;i<=k;i++)
        ans *= i;
    return ans;
}

int main() {
    ll n;
    while(scanf("%lld",&n) != EOF) {
        map <int,int> :: iterator iter;
        map <int,int> maps = prim_factor(n);
        int len = 0;
        for(iter=maps.begin();iter!=maps.end();iter++)
            len += iter->second;

        ll ans = mult(len);

        for(iter=maps.begin();iter!=maps.end();iter++)
            ans /= mult(iter->second);
        printf("%d %lld\n",len,ans);
    }
    return 0;
}