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[25];
void get_prime()    //埃氏筛法选取素数
{
    int i,j,p=0;
    memset(a,1,sizeof(a));
    a[0]=a[1]=0;
    for(i=2;i<=Max;i++)
    {
        if(a[i]==1)
        {
            prime[p++]=i;
            for(j=i*2;j<Max;j+=i)
                a[j]=0;
        }
    }
    init[1]=1;
    for(i=2;i<=20;i++)
        init[i]=i*init[i-1];
    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=0,k=0,t=1;
        i=0;
        while(n>1)
        {
            if(n%prime[i]==0)
            {
                u=0;
                while(n%prime[i]==0)
                {
                    n=n/prime[i];
                    u++;
                }
                t*=init[u];
                ans+=u;
                k++;
            }
            if(a[n]==1)
            {
                ans++;
                break;
            }
            i++;
        }
        p=init[ans];
        LL s=p/t;
        printf("%d %I64d
",ans,s);
    }
    return 0;
}