X-factor Chains（POJ3421 素数）

X-factor Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6212		Accepted: 1928

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
100=2*2*5*5;
所以最长为4，然后对2，2，5，5排列组合，但一直RE,TLE

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 #define LL long long
 #define Max ((1<<20)+2)
 bool a[Max];
 int prime[Max];
 int num[Max];
 LL init[];
 void get_prime()    //埃氏筛法选取素数
 {
     int i,j,p=;
     memset(a,,sizeof(a));
     a[]=a[]=;
     for(i=;i<=Max;i++)
     {
         if(a[i]==)
         {
             prime[p++]=i;
             for(j=i*;j<Max;j+=i)
                 a[j]=;
         }
     }
     init[]=;
     for(i=;i<=;i++)
         init[i]=i*init[i-];
     return;
 }
 int main()
 {
     int ans,n,i,j,k,u;
     get_prime();
     LL p,t;
     freopen("in.txt","r",stdin);
     while(scanf("%d",&n)!=EOF)
     {
         ans=,k=,t=;
         i=;
         while(n>)
         {
             if(n%prime[i]==)
             {
                 u=;
                 while(n%prime[i]==)
                 {
                     n=n/prime[i];
                     u++;
                 }
                 t*=init[u];
                 ans+=u;
                 k++;
             }
             if(a[n]==)
             {
                 ans++;
                 break;
             }
             i++;
         }
         p=init[ans];
         LL s=p/t;
         printf("%d %I64d\n",ans,s);
     }
     return ;
 }