1926: Factovisors N的阶乘的质因子个数

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最新推荐文章于 2022-03-18 11:25:01 发布

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分类专栏： joj acm_数学问题文章标签： integer input output each function less

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本文介绍了一种通过质因数分解判断一个数是否能整除另一个数的阶乘的算法，并提供了详细的实现代码。

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1926: Factovisors

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	659	122	Standard

The factorial function, n! is defined thus for n a non-negative integer:

   0! = 1
   n! = n * (n-1)!   (n > 0)

We say that a divides b if there exists an integer k such that

   k*a = b

The input to your program consists of several lines, each containing two non-negative integers, n and m, both less than 2^31. For each input line, output a line stating whether or not m divides n!, in the format shown below.

Sample Input

Output for Sample Input

9 divides 6!
27 does not divide 6!
10000 divides 20!
100000 does not divide 20!
1009 does not divide 1000!

对于任意质数p，n！中有（n/p+n/p^2+n/p^3+...)个质因子p。

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
int prime[50000],is[50000];
int pl;
int a[50000],cnt[50000];
//对于任意质数p，n！中有（n/p+n/p^2+n/p^3+...)个质因子p。
int countc(int n,int p)
{
    int ans=0;
    while(n/p)
    {
        ans+=n/p;
        n/=p;
    }
    return ans;
}
int main()
{
    for(int i=2;i<50000;i++)
    {
        if(is[i]==0) prime[pl++]=i;
        for(int j=0;j<pl&&prime[j]*i<50000;j++)
        {
            is[i*prime[j]]=1;
            if(i%prime[j]==0) break;
        }
    }
    int n,m;
    while(cin>>n>>m)
    {
        if(m==0) {printf("%d does not divide %d!/n",m,n);continue;}
        if(n>=m) {printf("%d divides %d!/n",m,n);continue;}
        int mm=m;
        double k=sqrt(m);
        int num=0;
        memset(cnt,0,sizeof(cnt));
        for(int i=0;prime[i]<=k;i++)
        {
            if(m%prime[i]==0)
            {
                a[num]=prime[i];
                while(m%prime[i]==0)
                {
                    cnt[num]++;
                    m/=prime[i];
                }
                num++;
            }
        }
        if(m>1) a[num]=m,cnt[num++]=1;
        int flag=1;
        for(int i=0;i<num;i++)
        {
            if(countc(n,a[i])<cnt[i])
            {
                flag=0;
                break;
            }
        }
        if(flag) printf("%d divides %d!/n",mm,n);
        else printf("%d does not divide %d!/n",mm,n);
    }
    return 0;
}