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, …, 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
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 ",ans,s); 64 } 65 return 0; 66 }