质因数计数公式 POJ 2992 Divisors

Divisors

http://poj.org/problem?id=2992

Time Limit:  1000MS

Memory Limit: 65536K

Description

Your task in this problem is to determine the number of divisors of  Cnk. Just for fun -- or do you need any special reason for such a useful computation?

Input

The input consists of several instances. Each instance consists of a single line containing two integers n and k (0 ≤ k ≤ n ≤ 431), separated by a single space.

Output

For each instance, output a line containing exactly one integer -- the number of distinct divisors of  Cnk. For the input instances, this number does not exceed 2 63 - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

题目大意：求c(n,m)的因子个数

思路：假设将一个数表示成它的质因数分解，如A=a^p1*b^p2*c^p3*...*n^pn.

那么它的约数个数就是：ans=(p1+1)*(p2+1)*(p3+1)*...*(pn+1).

而C(n,k)=n!/[(k!*(n-k)!]，c[n][k]代表n的阶乘时能够分解出几个k。

那么只需要求出他们的阶乘对于每一个素数的个数就可以了。

公式：ai=c[n][prime[i]]-c[k][prime[i]]-c[(n-k)][prime[i]]。ans=a1*a2.*...*ak (k代表当prime[k]小于n的时候)。

#include <stdio.h>
using namespace std;
const int maxn = 432;
bool vis[maxn];
int num[maxn][maxn];
int main()
{
	int n,k,i,j,t;
	//判断素数 
	for(i=2;i<22;i++)
	{
		if(!vis[i])
		{
			for(j=i*i;j<maxn;j+=i)
			vis[j]=true;
		}		
	}	
	//质因数计数公式		
	for(i=2;i<maxn;i++)
	{
		for(j=0;j<=i;j++)
		num[i][j]=num[i-1][j];
		t=i;
		for(j=2;j<maxn&&t>1;j++)
		{
			if(!vis[j])
			{
				while(t%j==0)
				{
					num[i][j]++;
					t/=j;
				}
			}				
		}			
	}
	while(~scanf("%d%d",&n,&k))
	{
		__int64 ans=1;
		if(k&&n-k)//k==0||(n-k)==0时，Cnk为1 
		{
			for(i=2;i<=n;i++)
			{
				if(!vis[i])
				ans*=num[n][i]-num[k][i]-num[n-k][i]+1;
			}				
		}			
		printf("%I64d\n",ans);
	}
	return 0;
}