LightOJ 1045

Factorial of an integer is defined by the following function

f(0) = 1
f(n) = f(n - 1) * n, if(n > 0)

So, factorial of 5 is 120. But in different bases, the factorial may be different. For example, factorial of 5 in base 8 is 170.

In this problem, you have to find the number of digit(s) of the factorial of an integer in a certain base.

Input

Input starts with an integer T (≤ 50000), denoting the number of test cases.

Each case begins with two integers n (0 ≤ n ≤ 106) and base (2 ≤ base ≤ 1000). Both of these integers will be given in decimal.

Output

For each case of input you have to print the case number and the digit(s) of factorial n in the given base.

Sample Input

5
5 10
8 10
22 3
1000000 2
0 100

Output for Sample Input

Case 1: 3
Case 2: 5
Case 3: 45
Case 4: 18488885
Case 5: 1

题目大意就是问N的阶乘在m进制下的位数。

对于这个题直接就是答案可能直接就出来了$int(log_m(n*(n-1)*(n-2)*(n-3)*...*2*1) + 1)$

但是由于测试数据样例过多，所以每次从1到n循环一遍代价会很大，所以直接打表1-1000000， 然后最后在除以$log(m)$

代码如下：

#include<bits/stdc++.h>
using namespace std;

vector<double> res(1000007);

void init()
{
    double sum = 0;
    for(int i=1; i<=1000000; ++ i)
    {
        sum += log(i);
        res[i] = sum;
    }
}

void solve(int cases)
{
    int n, m;
    scanf("%d%d", &n, &m);

    if(n == 0){
        printf("Case %d: %d
", cases, 1);
        return;
    }
    printf("Case %d: %d
", cases, int(res[n] / log(m) + 1));
}

int main()
{
    //ios::sync_with_stdio(false);

    init();

    int t;
    scanf("%d", &t);
    for(int i=1; i<=t; ++ i)
        solve(i);
    return 0;
}