POJ 3421

    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也是一样。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。

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代码（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;}

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

DescriptionGiven a positive integer X, an X-factor chain of length m is a sequence of integers,1 = X0, X1, X2, …, Xm = XsatisfyingXi < 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.InputThe input consists of several test cases. Each contains a positive integer X (X ≤ 220).OutputFor each test case, output the maximum length and the number of such X-factors chains.Sample Input23410100Sample Output1 11 12 12 24 6