UVA 10139 Factovisors（数论）

Factovisors

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

6 9
6 27
20 10000
20 100000
1000 1009

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!

题意：给出n和m，问m是否能整除n的阶乘。

分析：能够对m进行质因数分解，得到每一个素因子的个数。与n!中此因子的个数进行比較，若大于n!中此因子的个数。则不能整除。

#include<stdio.h>
#include<string.h>
#include<math.h>
const int MAXN = 100005;
int vis[MAXN], prime[10000], num;

void get_prime() //筛法求素数
{
    num = 0;
    memset(vis, 0, sizeof(vis));
    for(int i = 2; i < MAXN; i++)
    {
        if(!vis[i])
        {
            prime[num++] = i;
            for(int j = i + i; j < MAXN; j += i)
                vis[j] = 1;
        }
    }
}

int Cal(int w, int p) //计算w的阶乘中有多少个p
{
    int ans = 0;
    while(w)
    {
        w /= p;
        ans += w;
    }
    return ans;
}

bool judge(int n, int m)
{
	int k = (int)sqrt(m+0.5);
    for(int i = 0; i < num && prime[i] <= k; i++)
    {
        if(m % prime[i] == 0)
        {
            int cnt = 0;
            while(m % prime[i] == 0)
            {
                cnt++;
                m /= prime[i];
            }
            if(Cal(n, prime[i]) < cnt) return false;
        }
    } //此时若 m!=1，则m必为素数，假设n>=m。则m必然能够整除n!
    if(m > 1 && n < m) return false;
    return true;
}

int main()
{
    int n, m;
    get_prime();
    while(~scanf("%d%d",&n,&m))
    {
        if(judge(n, m)) printf("%d divides %d!\n", m, n);
        else printf("%d does not divide %d!\n", m, n);
    }
    return 0;
}

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