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.
6 9 6 27 20 10000 20 100000 1000 1009
9 divides 6! 27 does not divide 6! 10000 divides 20! 100000 does not divide 20! 1009 does not divide 1000!
#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，如果n>=m，则m必定可以整除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;
}UVA 10139 Factovisors（数论）,布布扣,bubuko.com
原文地址：http://blog.csdn.net/lyhvoyage/article/details/37911527
