Counting Factor Trees zoj 3405

Counting Factor Trees
Time Limit: 2 Seconds     
Memory Limit: 65536 KB
Factoring, i.e., listing all the prime factors, of an integer is a useful skill that often helps to solve math problems. For example, one of the ways to find the GCD (Greatest Common Divisor) or LCM (Least Common Multiple) of two integers is by listing all their prime factors. The GCD is then the product of all the common factors; the LCM is the product of all the remaining ones.
The Factor Tree is a tool for finding such prime factorizations. The figure below demonstrates three factor trees of 108. At the beginning a root with a number is given, say N, which is to be factored. Then, the root is factored into two children N1 and N2 such that N = N1 × N2 (N1 ≥ 2, N2 ≥ 2). Note that N1 and N2 need not be prime. The same factoring process continues until all the leaves are prime.
While the prime factorization is unique, the factor tree reflects the order in which the factors were found, and is by no means unique. So, how many factor trees of a number are there?

Input

There are no more than 10000 cases. A line containing an integer N (2 ≤ N ≤ 1000000000) is given for each case.

Output

Print the number of factor trees of N in a line for each case. The answer will be fit in a signed 64-bit integer.

Sample Input

12
108
642485760

Sample Output

6
140
9637611984000

Author: GAO, YuanContest: ZOJ Monthly, September 2010
비극적이다!시합 때 못해서 시합 후에 CE, TLE가 지나면 그만이고 플로팅 포인트 Error도 왔어요.
TLE의 Method 함수에 주의하십시오. 즉, x개의 요소가 얼마나 많은 배열 방법을 가진 함수를 구하는 것입니다.
원래 썼어요.
int Method(int num)
{
       if(a[num])
              return a[num];
       int i,j;
       int temp=0;
       for(i=1;i<=num/2;i++)
       {
              temp+=Method(i)*Method(num-i)*2;
       }
       if(num%2==0)
              temp-=Method(num/2)*Method(num/2);
       return temp;
}
 
= =!
이렇게 temp-=Method(num/2)*Method(num/2);다시 계산해야 되는 거 아니야!!
#include #include using namespace std; long long prime[5000]; long long factor[100]; long long r[100]; long long a[100]={0,1,1,2}; long long x; bool isprime(long long x) { long i,j; for(i=2;i<=sqrt(x*1.0);i++) { if(x%i==0) return false; } return true; } long long C(long long n,long long k) { long long tmp=1; if(k>(n-k)) k=n-k; for(long long i=1;i<=k;i++,n--) tmp=tmp*n/i; return tmp; } int cntnum=0; long long Divide(long long x) { long long i=1,j=0; cntnum=0; while(x!=1&&prime[i]*prime[i]<=x) { if(x%prime[i]==0) { factor[j]=prime[i]; r[j]=0; while(x%prime[i]==0) { x/=prime[i]; r[j]++; cntnum++; } j++; } i++; } if(x!=1) { cntnum++; r[j++]=1; }//cntnum->how many factors in total long long temp=1; long long tt=(long long)cntnum; for(i=0;i