LightOJ - 1045 Digits of Factorial （数学公式） 对数换低公式

LightOJ - 1045

Digits of Factorial

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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

Source

Problem Setter: Jane Alam Jan

//题意：输入n，m

表示给你一个数n让你求n的阶乘后的数再转换成m进制数的位数。

//思路：

运用对数换低公式log(a)(b)=log(c)(b)/log(c)(a);

#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;
}