poj 2649 Factovisors 对n!进行因数分解

Factovisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4431		Accepted: 1086

Description

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

Input

The input to your program consists of several lines, each containing two non-negative integers, n and m, both less than 2^31.

Output

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

Sample Output

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

Source

Waterloo local 1999.01.31

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
using namespace std;
const int MAXN=50000;
bool vis[MAXN+10];
int prime[MAXN+10];
int n,m,p;
vector<pair<int,int> > exp;

void init()//素数筛选
{
    memset(vis,0,sizeof(vis));
    p=0;
    for(int i=2; i<=MAXN; ++i)
    {
        if(vis[i]==false)
            prime[p++]=i;
        for(int j=0; j<p&&i*prime[j]<=MAXN; ++j)
            vis[i*prime[j]]=true;
    }
}

bool check(int x,int y)//判断当前的素数，n是否有这么多个
{
    int tmp=n,sum=0;
    while(tmp)
    {
        sum+=tmp/x;
        tmp/=x;
    }
    return y<=sum;
}

bool judge()
{
    for(int i=0; i<exp.size(); ++i)
        if(!check(exp[i].first,exp[i].second))
            return false;
    return true;
}

int main()
{
    init();
    while(scanf("%d%d",&n,&m)==2)
    {
        if(m==0)
        {
            printf("%d does not divide %d!
",m,n);
            continue;
        }
        exp.clear();
        int t=m;
        for(int i=0; i<p&&prime[i]<=m; ++i)
            if(m%prime[i]==0)
            {
                int cnt=0;
                while(m%prime[i]==0)
                {
                    m/=prime[i];
                    ++cnt;
                }
                exp.push_back(make_pair(prime[i],cnt));//m的素数个数与m的素数因子对应
            }
        if(m!=1)
            exp.push_back(make_pair(m,1));
        if(judge())
            printf("%d divides %d!
",t,n);
        else
            printf("%d does not divide %d!
",t,n);
    }
    return 0;
}