Factorial Problems 1

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 ≤ 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

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;
}
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