Divisors_组合数因子个数

Posted 2020-08-07 阿宝的锅锅

Description

Your task in this problem is to determine the number of divisors of C_n^k. 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 C_n^k. For the input instances, this number does not exceed 2⁶³ - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

 

 

 

 

【题意】求C(n,m)的质因子的个数。

【定理】设正整数n的所有素因子分解n=p1^a1*p2^a2*p3^a3****ps^as,那么T(n)=(a1+1)*(a2+1)*(a3+1)***(an+1);（求因子的个数的公式）

1.求出N以内素数

2.ei=[N/pi^1]+ [N/pi^2]+ …… + [N/pi^n] 其中[]为取整。即可以 int ei=0;while(N) ei+=(N/=pi);

3.套公式计算了，M=（e1+1）*（e2+1）*……*（en+1）

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=450;
int prime[1000]={2,3,5};
int k=3;
long long n,m,cnt[N][N];
void get_prime()//将1000以内的素数存入prime数组；
{
    int flag;
    int p=2;
    for(int i=7;i<=1000;i+=p)
    {
        flag=0;
        p=6-p;//巧妙的跳过了3的倍数，提高了效率
        for(int j=0;prime[j]*prime[j]<=i;j++)
        {
            if(i%prime[j]==0)
            {
                flag=1;
                break;
            }
        }
        if(!flag) prime[k++]=i;
    }
}
void init()
{
    memset(cnt,0,sizeof(cnt));
    get_prime();
    long long tmp,ret;
    for(int i=2;i<=431;i++)
    {
        for(int j=0;prime[j]<=i;j++)
        {
            tmp=i;
            ret=0;
            while(tmp)
            {
                tmp=tmp/prime[j];
                ret+=tmp;
            }
            cnt[i][prime[j]]=ret;//i的质因子数
        }
    }
}
int main()
{
    init();
    long long ret,ans;
    while(~scanf("%lld%lld",&n,&m))
    {
        ans=1;
        for(int i=0;prime[i]<=n;i++)
        {
            ret=cnt[n][prime[i]]-cnt[m][prime[i]]-cnt[n-m][prime[i]];//c(n,m)=n!/((n-m)!m!)，把对应因子个数相减，我们就得到了c(n,m)分解的结果
            ans*=(ret+1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}