**Description:**

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 ≤ 10 ^{6})** 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

# Sample Output:

Case 1: 3

Case 2: 5

Case 3: 45

Case 4: 18488885

Case 5: 1

………………………………………………

my solution:

#include <stdio.h> #include <iostream> #include <math.h> using namespace std; double a[1000005] = {0}; void cal () { for (int i = 1; i <= 1000005; i ++) { a[i] = a[i - 1] + log10(1.0 * i); } } int main () { int cas; int n, base; scanf("%d", &cas); cal(); for (int t = 1; t <= cas; t ++) { scanf("%d%d", &n, &base); printf("Case %d: %d\n", t, (int)(a[n] / log10(base * 1.0) + 1)); } return 0; }