uva 10780 分解质因数

想到分解质因数的话就很简单了，注意此题要求幂次至少为1，所以幂次为0即为Impossible to divide。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
using namespace std;

const int N = 10000;
bool visit[N];
int prime[N];
int n, m, ans, pn;

void sieve()
{
    memset( visit, 0, sizeof(visit) );
    visit[0] = visit[1] = 1;
    pn = 0;
    for ( int i = 0; i < N; i++ )
    {
        if ( !visit[i] )
        {
            prime[pn++] = i;
            for ( int j = i + i; j < N; j += i )
            {
                visit[j] = 1;
            }
        }
    }
}

int work( int num, int p )
{
    int res = 0;
    while ( num )
    {
        res += num / p;
        num /= p;
    }
    return res;
}

void solve()
{
    map<int, int> mp;
    int r = n;
    for ( int i = 0; prime[i] * prime[i] <= r; i++ )
    {
        if ( r % prime[i] == 0 )
        {
            while ( r % prime[i] == 0 )
            {
                mp[prime[i]]++;
                r = r / prime[i];
            }
        }
    }
    if ( r > 1 ) 
    {
        mp[r]++;
    }
    map<int, int>::iterator it;
    int ans = 9999999;    
    for ( it = mp.begin(); it != mp.end(); it++ )
    {
        int fir = (*it).first, sec = (*it).second;
        int cnt = work( m, fir );
        ans = min( ans, cnt / sec );
    }
    if ( !ans )
    {
        puts("Impossible to divide");
    }
    else
    {
        printf("%d
", ans);
    }
}

int main ()
{
    sieve();
    int t;
    scanf("%d", &t);
    for ( int _case = 1; _case <= t; _case++ )
    {
        printf("Case %d:
", _case);
        scanf("%d%d", &n, &m);
        solve();
    }
    return 0;
}