2992 Divisors 求C(N,K)的因子个数_求c(n,k)的不同因子个数。-优快云博客

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/5910542

本文介绍了一种计算组合数 C(n, k) 的不同除数数量的方法。通过预先计算素数并利用这些素数来确定给定组合数的素因子个数，进而计算出该组合数的除数总数。文章提供了完整的 C++ 实现代码。

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Divisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6785		Accepted: 1781

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 2 ⁶³ - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

n!的素因子i的个数 = (n-1)!的素因子i的个数 + n的素因子i的个数
c(n,k)的素因子i的个数 = n!的素因子i的个数- (n-k)!的素因子i的个数 - k!的素因子i的个数

#include<iostream>
#include<cstdio>
using namespace std;
int is[500],prime[500],pl;
void _prime()
{
    for(int i=2;i<500;i++)
    {
        if(is[i]==0) prime[pl++]=i;
        for(int j=0;j<pl&&i*prime[j]<500;j++)
        {
            is[i*prime[j]]=1;
            if(i%prime[j]==0) break;
        }
    }
}
__int64 _count(int n,int p)
{
    __int64 cnt=0;
    while(n)
    {
        cnt+=n/p;
        n/=p;
    }
    return cnt;
}
int main()
{
    _prime();
    int n,k;
    while(scanf("%d%d",&n,&k)==2)
    {
        __int64 cnt=1;
        for(int i=0;prime[i]<=n;i++)
        {
            cnt*=(1+_count(n,prime[i])-_count(k,prime[i])-_count(n-k,prime[i]));
        }
        printf("%I64d/n",cnt);
    }
    return 0;
}