TJU2848(Factorial )

 

2845.   Factorial

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Time Limit: 1.0 Seconds    Memory Limit: 65536K
Total Runs: 1408    Accepted Runs: 411

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Robby is a clever boy, he can do multiplication very quickly, even in base-2 (binary system), base-16 (hexadecimal system) or base-100000.

Now he wants to challenge your computer, the task is quite simple: Given a positive integer N, if we express the N ! (N ! = N * (N - 1) * (N - 2) * ... * 2 * 1) in base-B, how many ZEROs there will be at the end of the number. Can you make a wonderful program to beat him?

Input

Each test case contain one line with two numbers N and B. (1 ≤ N ≤ 10⁹, 2 ≤ B ≤ 100000)

The input is terminated with N = B = 0.

Output

Output one line for each test case, indicating the number of ZEROs.

Sample Input

7 10

7 10000

7 2

0 0

 

Sample Output

1

0

4

 

Hint

7! = (5040)₁₀, so there will be one zero at the end in base-10. But in base-10000, the number (5040)₁₀ can be expressed in one non-zero digit, so the second answer is 0. In base-2, 7! = (1001110110000)₂, so the third answer is 4.

Problem Setter: RoBa

/**/ /*
    2009-05-11 14:07:04 Accepted 2845 C++ 0.8K 0'00.00" 1204K 
*/
#include <iostream>
using namespace std;

int get_ans(int n, int b)
{
    int sum = 0;

    while(n / b)
    {
        sum += n / b;
        n /= b;
    }
    return sum;
}

int get_small_ans(int n, int b)
{
    bool flag = true;
    int cnt, ans = 0;
    int i;

    for(i = 2; i * i <= b; i++)
    {
        cnt = 0;
        while(b % i == 0)
        {//每次去进制的质因子
            b /= i;
            cnt++;
        }

        if(cnt)
        {
            if(flag)
            {
                ans = get_ans(n, i) / cnt;
                flag = false;
            }
            else //取进制质因子中的最小一个
                ans = min(ans, get_ans(n, i) / cnt);
        }

    }

    if(b > 1)
    {    
        if(flag)
            ans = get_ans(n, b);
        else
            ans = min(ans, get_ans(n, b));
    }
    return ans;
}

int main()
{
    int n, b;

    while(scanf("%d%d", &n, &b) != EOF)
    {
        if(!n && !b)
            break;
        printf("%d\n", get_small_ans(n, b));
    }
    return 0;
}