X-factor Chains（POJ3421 素数）-优快云博客

本文探讨了X-factorChains问题的求解方法，通过分解正整数X为素数因子，利用埃氏筛法寻找所有可能的素数，并计算X-factorChains的最大长度及相应链条的数量。

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

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

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 #define LL long long
 6 #define Max ((1<<20)+2)
 7 bool a[Max];
 8 int prime[Max];
 9 int num[Max];
10 LL init[25];
11 void get_prime()    //埃氏筛法选取素数
12 {
13     int i,j,p=0;
14     memset(a,1,sizeof(a));
15     a[0]=a[1]=0;
16     for(i=2;i<=Max;i++)
17     {
18         if(a[i]==1)
19         {
20             prime[p++]=i;
21             for(j=i*2;j<Max;j+=i)
22                 a[j]=0;
23         }
24     }
25     init[1]=1;
26     for(i=2;i<=20;i++)
27         init[i]=i*init[i-1];
28     return;
29 }
30 int main()
31 {
32     int ans,n,i,j,k,u;
33     get_prime();
34     LL p,t;
35     freopen("in.txt","r",stdin);
36     while(scanf("%d",&n)!=EOF)
37     {
38         ans=0,k=0,t=1;
39         i=0;
40         while(n>1)
41         {
42             if(n%prime[i]==0)
43             {
44                 u=0;
45                 while(n%prime[i]==0)
46                 {
47                     n=n/prime[i];
48                     u++;
49                 }
50                 t*=init[u];
51                 ans+=u;
52                 k++;
53             }
54             if(a[n]==1)
55             {
56                 ans++;
57                 break;
58             }
59             i++;
60         }
61         p=init[ans];
62         LL s=p/t;
63         printf("%d %I64d\n",ans,s);
64     }
65     return 0;
66 }