LightOJ-1045 Digits of factorial (mathematical formula) logarithmic low formula

Source: Internet
Author: User
LightOJ-1045 Digits of factorial
Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %lld &%llu

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Description

Factorial of an integer was 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 is 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 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 is begins with the integers n (0≤n≤106) and Base (2≤base≤1000). Both of these integers would be a given in decimal.

Output

For each case of input, you had 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 4:18,488,885

Case 5:1

Source problem Setter:jane Alam Jan//Test instructions: Enter n,m to give you a number n allows you to convert the number of the factorial of N to the number of M-binary digits. Idea: Log (a) (b) =log (c) (b)/log (c) (a) using logarithmic commutation formula

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#define N 1000010
using namespace std;
Double sum[n];
int init ()
{for
	(int i=1;i<=n;i++)
		sum[i]=sum[i-1]+log10 (i);
}
int main ()
{
	int t,t=1,n,m;
	Init ();
	scanf ("%d", &t);
	while (t--)
	{
		scanf ("%d%d", &n,&m);
		printf ("Case%d:%d\n", t++, (int) (SUM[N]/LOG10 (m)) +1);
	}
	return 0;
}

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