Again Prime? No time.
Input: standard input
Output: standard output
Time Limit: 1 second

The problem statement is very easy. Given a number n you have to determine the largest power of m, not necessarily prime, that divides n!.

Input

The input file consists of several test cases. The first line in the file is the number of cases to handle. The following lines are the cases each of which contains two integers m (1<m<5000)

For each case in the input, print the case number and result in separate lines. The result is either an integer if mdivides n! or a line "Impossible to divide

Sample Input

2
2 10
2 100

Sample Output

Case 1:
8
Case 2:
97

題意：給你m,n求最大的k使得是n！的因子；

思路：把m質因子分解為，設p為pi中最大的素數，,然後求n!中p的個數即為為b，即為答案（對於i，求出相應的b，最小的）

程式碼：

<pre name="code" class="cpp">#include <iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<stdlib.h>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<set>
#define inf 0x3f3f3f3f
#define eps 1e-5
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
using namespace std;
int sum(int n,int p)
{
    int x=p,ans=0;
    while(n/p!=0)
    {
        ans+=n/p;
        p*=x;
    }
    return ans;
}
int main()
{
    int t,n,m,i,j;
    scanf("%d",&t);
    {
        for(i=1;i<=t;i++)
        {
            scanf("%d%d",&m,&n);
            printf("Case %d:\n",i);
            int ans=inf,pnum;
            for(j=2;m>1;j++)
            {
                pnum=0;
                while(m%j==0)
                {
                    pnum++;
                    m/=j;
                }
                if(pnum)
                {
                    int tt=sum(n,j)/pnum;
                    if(ans>tt)ans=tt;
                }
            }
            if(ans)
                printf("%d\n",ans);
            else
                printf("Impossible to divide\n");
        }
    }
    return 0;
}