（zju3405）Counting Factor Trees

Counting Factor Trees

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

12108642485760

Sample Output

61409637611984000

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Author: GAO, Yuan
Contest: ZOJ Monthly, September 2010
题意：给出一个数，然后统计其质因子组成的完全二叉树有多少种。

首先对于n个叶子的完全二叉树 其个数为卡塔兰数序列的第n-1个值（求卡塔兰数可用状态转移方程 dp[i] = sum(dp[j] * dp[i - 1 - j] | 0 <= j < i)）.

然后对其质因子进行排列，C(sum,a) * C(sum - a.b) * C(sum - a - b,c) ……；

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <cmath>using namespace std;typedef struct Node{int factor;int cnt;Node(){cnt = 0;}}node;node ans[1001];unsigned long long dp[1002];unsigned long long sum;void sep(int n) //分解{int i;int s = sqrt(1.0 * n + 1);for(i = 2; i <= s; i++){if(n % i == 0){ans[sum].factor = i;ans[sum].cnt = 1;n /= i;while(n % i == 0){n /= i;ans[sum].cnt++;}sum++;s = sqrt(1.0 * n + 1);}}if(n > 1){ans[sum].factor = n;ans[sum++].cnt = 1;;}}void cant(){dp[0] = 1;int i,j;for(i = 1; i <= 65; i++){for(j = 0; j < i; j++){dp[i] += dp[j] * dp[i - j - 1];}}}unsigned long long per(int n,int m){int i;unsigned long long s = 1;for(i = 0; i < m; i++){s = s * (n - i) / (i + 1);}return s;}int main(){int n;cant();int i;while(scanf("%d",&n)!=EOF){sum = 0;sep(n);int Count = 0;for(i = 0; i < sum; i++){Count += ans[i].cnt;}long long ret = dp[Count - 1];for(i = 0; i < sum; i++){ret *= per(Count,ans[i].cnt);Count -= ans[i].cnt;}printf("%llu/n",ret);}return 0;}