Lightoj1045 Digits of Factorial(数论)

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

把n!(十进制)转化为base进制有多少位

取位数用log  但是直接取会爆  可以把它拆开

log(n!) = log(n)+log(n-1) + log(n-2) +...+log(2)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
double dp[1000001];
int main()
{
	int t,n,base;
	scanf("%d",&t);
	dp[0]=0;
	for(int i=1;i<1000001;i++)
	{
	 dp[i]+=dp[i-1]+log(i);	
	}
	for(int i=1;i<=t;i++)
	{
		scanf("%d%d",&n,&base);
		printf("Case %d: %d
",i,int(dp[n]/log(base))+1);
	}
	return 0;
}