X-factor Chains（素数）

原题链接X-factor Chains
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6575 Accepted: 2054
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
Source

POJ Monthly–2007.10.06, ailyanlu@zsu

//http://poj.org/problem?id=3421#include <algorithm>#include <iostream>#include <utility>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>using namespace std;typedef long long ll;typedef unsigned long long ull;const int MOD = int(1e9) + 7;//int MOD = 99990001;const int INF = 0x3f3f3f3f;const ll INFF = (~(0ULL)>>1);const double EPS = 1e-9;const double OO = 1e20;const double PI = acos(-1.0); //M_PI;const int fx[] = {-1, 1, 0, 0};const int fy[] = {0, 0, -1, 1};const int maxn=10000 + 5;int n;vector<int> res;//得到一个数的阶乘ull f(int x){        ull ans=1;        for(int i=2;i<=x;i++)                ans*=i;        return ans;}//得到一个数的所有质因数的幂void solve(int x){        for(int i=2;i*i<=x;i++){                int t=0;//因为从低位开始，所以一定所有的进入循环的i都是质因子                while(x%i==0) {t++;x/=i;}                res.push_back(t);        }        if(x!=1) res.push_back(1);        return;}int main(){        while(scanf("%d",&n)==1){                res.clear();                solve(n);                int len=res.size();                int sum=0;//sum代表所有质因子的幂的和                for(int i=0;i<len;i++)                        sum+=res[i];                ull ans=f(sum);//得到sum的阶乘                for(int i=0;i<len;i++)                        ans/=f(res[i]);//除以其中一个质因数的幂的阶乘                cout << sum << " " << ans << endl;        }        return 0;}/*因子链：将一个数X分解成从1到X的数列，前一个数可以整除后一个数，求最大链长和链的个数。因为题目里要求Xi | Xi+1，而Xm又限定为X，所以我们可以想到Xm-1是X除以其某个约数得到的，Xm-1也是可以由Xm-2的到的，可以认为这些单位之间就是插入了一个乘数。由此我们可以知道“X-factor Chains”是通过不断乘以X的约数得到的，为了长度最大，所以约数必须是素数。通过记录有哪些素因数，以及素因子的数量，我们就可以得到链的长度。而长度最大的链的数量则是这些数的排列数，公式为素因子个数之和的阶乘/每个素因子个数的阶乘。*/